Given
Wholesale price for a pair of shoes = $3.50.
Wholesale price by 60%.
Find
price of the pair of shoes in the department store
Explanation
Price in deparrtmental store is 60% more than wholesale
[tex](1+60\%\text{ \rparen of wholesale }[/tex]so ,
[tex]\begin{gathered} 1.6\times3.50 \\ 5.60 \end{gathered}[/tex]Final Answer
Therefore , the price of the pair of shoes in the department store is $5.60.
The table shows the fraction of students from differentgrade levels who are in favor of adding new items tothe lunch menu at their school. Which list shows the grade levels in order from the greatest fraction of students to the least fraction of students ?
First, write all the fractions using the same denominator. To do so, find the least common multiple of all denominatos. The denominators are:
[tex]50,20,25,75,5[/tex]The least common multiple of all those numbers is 300.
Use 300 as a common denominator for all fractions to be able to compare their values.
5th grade
[tex]\frac{33}{50}=\frac{33\times6}{50\times6}=\frac{198}{300}[/tex]6th grade
[tex]\frac{13}{20}=\frac{13\times15}{20\times15}=\frac{195}{300}[/tex]7th grade
[tex]\frac{18}{25}=\frac{18\times12}{25\times12}=\frac{216}{300}[/tex]8th grade
[tex]\frac{51}{75}=\frac{51\times4}{75\times4}=\frac{204}{300}[/tex]9th grade
[tex]\frac{3}{5}=\frac{3\times60}{5\times60}=\frac{180}{300}[/tex]Now, we can compare the numerators to list the fraction from greatest to lowest:
[tex]\begin{gathered} \frac{216}{300}>\frac{204}{300}>\frac{198}{300}>\frac{195}{300}>\frac{180}{300} \\ \Leftrightarrow\frac{18}{25}>\frac{51}{75}>\frac{33}{50}>\frac{13}{20}>\frac{3}{5} \\ \Leftrightarrow7th\text{ grade}>8th\text{ grade}>5th\text{ grade}>6th\text{ grade}>9th\text{ grade} \end{gathered}[/tex]Therefore, the list of grade levels in order from the greatest fraction of students to the least fraction of students, is:
7th grade (18/25)
8th grade (51/75)
5th grade (33/50)
6th grade (13/20)
9th grade (3/5)
What is the solution to the equation below? 6x= x + 20 O A. x = 4 B. X = 20 C. x = 5 D. No Solutions
Simplify the equation 6x = x +20 to obtain the value of x.
[tex]\begin{gathered} 6x=x+20 \\ 6x-x=20 \\ 5x=20 \\ x=\frac{20}{5} \\ =4 \end{gathered}[/tex]So answer is x = 4
Option A is correct.
Describe the relationship between the number 2 x 10^4 and 4 x 10^6
2 * 10^4 = 2 * 10000 = 20,000
4 * 10^6 = 4 * 1000000 = 4,000,000
4,000,000/20,000 = 200
Therefore, 4 * 10^6 is 200 times 2 * 10^4
use the formula Sn to find the sum of the first five terms of the geometric sequence.
First, find the common ratio r:
-4/9 : 4/3 = -1/3
4/3 : -4 = -1/3
-4:12 = -1/3
r= -1/3
[tex]Sn=\frac{a(r^n-1)}{r-1}[/tex]Where:
a= first term = 12
n= number of terms = 5
Replacing:
[tex]Sn=\frac{12(-\frac{1^{}}{3}^5-1)}{-\frac{1}{3}-1}[/tex]Sn= 244/27 = 9 1/27
Find the volume of the given solid.Round to the nearest 10th, If necessary. In cubic inches
ANSWER
33.5 cubic inches
EXPLANATION
This is a cone with radius r = 2 in and height h = 8 in. The volume of a cone is,
[tex]V=\frac{1}{3}\cdot\pi\cdot r^2\cdot h[/tex]Replace the known values and solve,
[tex]V=\frac{1}{3}\cdot\pi\cdot2^2in^2\cdot8in=\frac{32}{3}\pi\text{ }in^3\approx33.5\text{ }in^3[/tex]Hence, the volume of the cone is 33.5 in³, rounded to the nearest tenth.
identify point in region of inequalities
We want to picture the inequalities
[tex]y<\text{ - x -3}[/tex]and
[tex]y>\frac{4}{5}x\text{ +5}[/tex]First, we consider the lines y= -x -3 and and y=(4/5) x +5 . Since the first line has a negative slope, this means that its graph should go downwards as x increases and since the other line has a positive slope, this means that its graph should go upwards as x increases. This leads to the following picture
Now, the expression
[tex]y<\text{ -x -3}[/tex]means that the y coordinate of the line should be below the red line. Also, the expression
[tex]y>\frac{4}{5}x+5[/tex]means tha the y coordinate should be above the blue line. If we combine both conditions, we find the following region
so we should look for a point that lies in this region
We are given the points (-1,9), (-6,2), (9,-9) and (-8,-5).
We see that the yellow region is located where the x coordinate is always negative. So, this means that we discard (9,-9).
so we should test the other points. Since -8 is the furthest to the left, let us calculate the value of each line at x=-8.
[tex]\text{ -(-8) -3 = 8 -3 = 5}[/tex]so, in this case the first expression is accomplished since -5 < 5. And
[tex]\frac{4}{5}\cdot(\text{ -8)+5= =}\frac{\text{ -7}}{5}=\text{ -1.4}[/tex]However note that -5 < 1.4, and it should be greater than -1.4 to be in the yellow region. So we discard the point (-8,-5) .
We can check , iusing the graph, that the lines cross at the point (-40/9, 13/9) which is about (-4.44, 1.44). This means that for the point to be on the yellow region, it should be on the left of -4.44. Since the only point that we are given that fulfills this condition is (-6, 2), this should be our answer. We check that
[tex]\text{ -(-6)-3=3>2}[/tex]and
[tex]\frac{4}{5}\cdot(\text{ -6)+5 = }\frac{1}{5}=0.2<2[/tex]so, the point (-6,2) is in the yellow region
Which points sre vertices of the pre-image, rectangle ABCD?Makes no sense
Given rectangle A'B'C'D', you know that it was obtained after translating rectangle ABCD using this rule:
[tex]T_{-4,3}(x,y)[/tex]That indicates that each point of rectangle ABCD was translating 4 units to the left and 3 units up, in order to obtain rectangle A'B'C'D'.
Notice that the coordinates of the vertices of rectangle A'B'C'D' are:
[tex]\begin{gathered} A^{\prime}(-5,4) \\ B^{\prime}(3,4) \\ C^{\prime}(3,1) \\ D^{\prime}(-5,1) \end{gathered}[/tex]Therefore, in order to find the coordinates of ABCD, you can add 4 units to the x-coordinate of each point and subtract 3 units to each y-coordinate of each point. You get:
[tex]\begin{gathered} A=(-5+4,4-3)=(-1,1) \\ B=^(3+4,4-3)=(7,1) \\ C=(3+4,1-3)=(7,-2) \\ D=(-5+4,1-3)=(-1,-2) \end{gathered}[/tex]Hence, the answers are:
- First option.
- Second option.
- Fourth option.
- Fifth option.
After a translation, the image of P(-3, 5) is P'(-4, 3). Identify the image of the point (1, 6) after this same translation.
The image of the point (1, 6) after the translation is (0, 4).
What is named as translation?In geometry, translation refers to a function that shifts an object a specified distance. The object is not elsewhere altered. It has not been rotated, mirrored, or resized.Every location of the object should be relocated in the same manner and at the same distance during a translation.When performing a translation, this same initial object is referred to as the pre-image, as well as the object that after translation is referred to as the image.For the given question,
The image of point P(-3, 5) after a translation is P'(-4, 3).
In this, there is a shift of 1 units to the left of x axis and shift of 2 units up on the y axis.
Thus, do the same translation for the point (1, 6).
After translation image will be (0, 4)
Thus, image of the point (1, 6) after the translation is (0, 4).
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si f(x) = x + 5 cuanto es f(2) f(1) f(0) f(-1) f-(-2) f(a)
f (x)= x+ 5
f(2)
Reemplaza x por 2 y resuelve
f(2)= 2 + 5 = 7
Mismo procedimiento para los demas valores:
f(1) = 1 + 5 = 6
f(0) = 0 + 5 = 5
f(-1)= -1+5 = 4
f(-2)= -2+5 = 3
f(a)= a + 5
**Line m is represented by the equation -2x + 4y = 16. Line m and line k are Blank #1:
Line m:
[tex]y=\frac{2}{3}x+4[/tex]line k:
[tex]\begin{gathered} -2x+4y=16 \\ 4y=2x+16 \\ y=\frac{2x+16}{4} \\ y=\frac{x}{2}+4 \end{gathered}[/tex]so, the lime m and line k are:
D. Neither parallel nor perpendicular
Because:
D. their slopes have no relationship
Owners of a recreation area are filling a small pond with water. They are adding water at a rate of 29 L per minute. There are 400 L in the pond to start. Let W represent the total amount of water in the pond (in liters) and let T represent the total number of minutes that water has been added.Write an equation relating W to T. Then use this equation to find the total amount of water after 13 minutes.Equation : Total amount of water after 13 minutes : liters
In this problem, we have a linear equation of the form
W=mT+b ----> equation in slope-intercept form
where
m is the unit rate or slope of the linear equation
m=29 L/min ----> given
b is the initial value
b=400 L ----> given
substitute
W=29T+400 -------> equation relating W to T.For T=13 min
substitute
W=29(13)+400
W=777 L
the total amount of water after 13 minutes is 777 LBecause of damage, a computer company had 5 tablets returned out of the 80 that were sold. Suppose the number of damaged tablets sold continue at this rate. How many tablets should the company expect to have returned if it sells 400 of them?
we are told that there 5 damaged tablets out of 80 that are sold. Therefore, the rate of damaged tablets per sold tablets is:
[tex]\frac{5\text{ damaged}}{80\text{ sold}}[/tex]Multiplying this rate by the 400 sold tablets we get:
[tex]\frac{5\text{ damaged}}{80\text{ sold}}\times40\text{0 sold}[/tex]Solving we get:
[tex]\frac{5\text{ damaged}}{80\text{ sold}}\times40\text{0 sold}=25\text{ damaged}[/tex]Therefore, if the rate continues, the company can expect to return 25 tablets.
1) The perimeter of a rectangular garden is 344M. If the width of the garden is 76M, what is its length?
Equation:
Solution:
(I need the equation and solution)
2) The area of a rectangular window is 7315CM^2 (^2 is squared). If the length of the window is 95CM, what is its width?
Equation:
Solution:
(Once again, I need the equation and solution)
3) The perimeter of a rectangular garden is 5/8 mile. If the width of the garden is 3/16 mile, what is its length?
4) The area of a rectangular window is 8256M^2 (^2 is squared). If the length of the window is 86M, what is its width?
5) The length of a rectangle is six times its width. The perimeter of the rectangle is 98M, find its length and width.
6) The perimeter of the pentagon below is 58 units. Find VW. Write your answer without variables.
The length of the rectangle is 96 m, the width of the rectangle is 77 cm , the length of the rectangle is 1/8 mile, the length and width of the rectangle is 7 m and 42 m respectively, VW is 11 units.
According to the question,
1) Perimeter of rectangle = 344 M
Width = 76 M
Perimeter of rectangle = 2(length + width)
2(length+76) = 344
length+76 = 172
length = 172-76
Length of the rectangle = 96 M
2) Area of a rectangular window = 7315 [tex]cm^{2}[/tex]
Length of the window is 95 cm.
Area of rectangle = length*width
95*width = 7315
width = 7315/95
Width of the rectangle = 77 cm
3) The perimeter of a rectangular garden is 5/8 mile.
The width of the garden is 3/16 mile.
Perimeter of rectangle = 2(length+width)
2(length+3/16) = 5/8
length+3/16 = 5/(2*8)
length = 5/16-3/16
Length of the rectangle = 2/16 or 1/8 mile
4) The area of a rectangular window is 8256 [tex]m^{2}[/tex].
The length of the window is 86 m.
Area of rectangle = length*width
86*width = 8256
width = 8256/86
Width of the rectangle = 96 m
5) The length of a rectangle is six times its width. The perimeter of the rectangle is 98 m.
Let's take width of the rectangle to be x m.
Length of rectangle = 6x m
2(length+width) = 98
2(6x+x) = 98
2*7x = 98
14x = 98
x = 98/14
x = 7 m
Width = 7 m
Length = 7*6 m or 42 m
6) The perimeter of the pentagon is 58 units.
3z+10+z+3+2z-1+10 = 58
6z+10+3-1+10 = 58
6z+22 = 58
6z = 58-22
6z = 36
z = 36/6
z = 6 units
VW = 2z-1
VW = 2*6-1
VW = 12-1
VW = 11 units
Hence, the answer to 1 is 96 m , 2 is 77 cm, 3 is 1/8 mile , 4 is 96 m , 5 is 7 m and 42 m and 6 is 11 units.
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100 Points
A rectangle has sides measuring (2x + 5) units and (3x + 7) units.
Part A: What is the expression that represents the area of the rectangle? Show your work.
Part B: What are the degrees and classifications of the expression obtained in Part A?
Part C: How does Part A demonstrate the closure property for the multiplication of polynomials?
The expression that represents the area of the rectangle is 6x² + 29x + 35. The degree of the expression will be 2. And the closure property of multiplication is also demonstrated.
What is the area of the rectangle?Let W be the rectangle's width and L its length.
Area of the rectangle = L × W square units
The sides of a rectangle are (2x + 5) units and (3x + 7) units, respectively. Then the area of the rectangle will be given as,
A = (2x + 5)(3x + 7)
A = 2x(3x + 7) + 5(3x + 7)
A = 6x² + 14x + 15x + 35
A = 6x² + 29x + 35
The degree of the expression will be 2. And the closure property of multiplication is also demonstrated.
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Answer:
[tex]\textsf{A.} \quad \textsf{Area}=(2x+5)(3x+7)[/tex]
B. Degree = 2.
Classification = Quadratic trinomial.
C. Part A demonstrates the closure property for the multiplication of polynomials as the multiplication of the two given polynomials (side measures) produces another polynomial (area).
Step-by-step explanation:
Part AArea of a rectangle
[tex]\boxed{A=lw}[/tex]
where l is the length and w is the width.
Given that a rectangle has sides measuring (2x + 5) units and (3x + 7) units, the area can be expressed as a product of the two sides:
[tex]\implies \textsf{Area}=(2x+5)(3x+7)[/tex]
Part BFOIL method
[tex]\boxed{(a + b)(c + d) = ac + ad + bc + bd}[/tex]
Expand the brackets of the equation found in part A by using the FOIL method:
[tex]\implies \textsf{Area}=6x^2+14x+15x+35[/tex]
[tex]\implies \textsf{Area}=6x^2+29x+35[/tex]
The degree of a polynomial is the highest power of a variable in the polynomial equation. Therefore:
The degree of the function is 2.A polynomial is classified according to the number of terms and its degree.
The number of terms in the polynomial is three, therefore it is a trinomial.The degree of the function is 2, therefore it is quadratic.Part CClosure property under Multiplication
A set is closed under multiplication when we perform that operation on elements of the set and the answer is also in the set.
Therefore, Part A demonstrates the closure property for the multiplication of polynomials as the multiplication of the two given polynomials (side measures) produces another polynomial (area).
The odds in favor of a horse winning a race are 7:4. Find the probability that the horse will win the race.A. 7/12B. 4/7C. 7/11D. 4/11
We have a reason for 7:4,
i.e. the total probability of winning is 7+4=11
If the horse has a probability of winning of 7 between 11
We can say that the Pw of the horse is as follows
[tex]\frac{7}{11}[/tex]The answer is the option C
Consider the quadratic f(x)=x^2-x-30Determine the following ( enter all numerical answers as integers,fraction or decimals$The smallest (leftmost) x-intercepts is x=The largest (rightmost)x-intercepts is x=The y-intercept is y=The vertex is The line of symmetry has the equation
ANSWER
Smallest x-intercept: x = -5
Largest x-intercept: x = 6
y-intercept: y = -30
The vertex is (1/2, -121/4)
Line of symmetry x = 1/2
EXPLANATION
Given:
[tex]f(x)\text{ = x}^2\text{ - x - 30}[/tex]Desired Results:
1. Smallest x-intercept: x =
2. Largest x-intercept: x =
3. y-intercept: y =
4. The vertex is
5. Equation of Line of symmetry
1. Determine the x-intercepts by equating f(x) to zero (0).
[tex]\begin{gathered} 0\text{ = x}^2-x-30 \\ x^2-6x+5x-30\text{ = 0} \\ x(x-6)+5(x-6)=0 \\ (x-6)(x+5)=0 \\ x-6=0,\text{ x+5=0} \\ x\text{ = 6, x = -5} \end{gathered}[/tex]The smallest and largest x-intercepts are -5 and 6 respectively.
2. Determine the y-intercept by equating x to 0
[tex]\begin{gathered} y\text{ = \lparen0\rparen}^2-0-30 \\ y\text{ = -30} \end{gathered}[/tex]y-intercept is -30
3a. Determine the x-coordinate of the vertex using the formula
[tex]x\text{ = -}\frac{b}{2a}[/tex]where:
a = 1
b = -1
Substitute the values
[tex]\begin{gathered} x\text{ = -}\frac{(-1)}{2(1)} \\ x\text{ = }\frac{1}{2} \end{gathered}[/tex]3b. Determine the y-coordinate of the vertex by substituting x into the equation
[tex]\begin{gathered} y\text{ = \lparen}\frac{1}{2})^2-\frac{1}{2}-30 \\ y\text{ = }\frac{1}{4}-\frac{1}{2}-30 \\ Find\text{ LCM} \\ y\text{ = }\frac{1-2-120}{4} \\ y\text{ = -}\frac{121}{4} \end{gathered}[/tex]4. Determine the line of symmetry
In standard form the line of symmetry of a quadratic function can be identified using the formula
[tex]\begin{gathered} x\text{ = -}\frac{b}{2a} \\ x\text{ = }\frac{1}{2} \end{gathered}[/tex]Kiran is solving 2x-3/x-1=2/x(x-1) for x, and he uses these steps.He checks his answer and finds that it isn’t a solution to the original equation, so he writes “no solutions.” Unfortunately, Kiran made a mistake while solving. Find his error and calculate the actual solution(s).
Solution:
Given:
[tex]\begin{gathered} To\text{ solve,} \\ \frac{2x-3}{x-1}=\frac{2}{x(x-1)} \end{gathered}[/tex]Kiran multiplied the left-hand side of the equation by (x-1) and multiplied the right-hand side of the equation by x(x-1).
That was where he made the mistake. He ought to have multiplied both sides with the same quantity (Lowest Common Denominator) so as not to change the actual value of the question.
Multiplying both sides by the same quantity does not change the real magnitude of the question.
The actual solution goes thus,
[tex]\begin{gathered} \frac{2x-3}{x-1}=\frac{2}{x(x-1)} \\ \text{Multiplying both sides of the equation by the LCD,} \\ \text{The LCD is x(x-1)} \\ x(x-1)(\frac{2x-3}{x-1})=x(x-1)(\frac{2}{x(x-1)}) \\ x(2x-3)=2 \\ \text{Expanding the bracket,} \\ 2x^2-3x=2 \\ \text{Collecting all the terms to one side to make it a quadratic equation,} \\ 2x^2-3x-2=0 \end{gathered}[/tex]Solving the quadratic equation;
[tex]\begin{gathered} 2x^2-3x-2=0 \\ 2x^2-4x+x-2=0 \\ \text{Factorizing the equation,} \\ 2x(x-2)+1(x-2)=0 \\ (2x+1)(x-2)=0 \\ 2x+1=0 \\ 2x=0-1 \\ 2x=-1 \\ \text{Dividing both sides by 2,} \\ x=-\frac{1}{2} \\ \\ \\ OR \\ x-2=0 \\ x=0+2 \\ x=2 \end{gathered}[/tex]Therefore, the actual solutions to the expression are;
[tex]\begin{gathered} x=-\frac{1}{2} \\ \\ OR \\ \\ x=2 \end{gathered}[/tex]A survey of 100 high school students provided thisfrequency table on how students get to school:Drive toTake theGradeWalkSchoolbusSophomore2253Junior13202Senior2555Find the probability that a randomly selected studenteither takes the bus or walks.[?P(Take the bus U Walk)
Let's call the event of a student taking the bus as event A, and the event of a student walking as event B. The theoretical probability is defined as the ratio of the number of favourable outcomes to the number of possible outcomes. We have a total of 100 students, where 50 of them take the bus and 10 of them walk. This gives to us the following informations:
[tex]\begin{gathered} P(A)=\frac{50}{100} \\ P(B)=\frac{10}{100} \end{gathered}[/tex]The additive property of probability tells us that:
[tex]P(A\:or\:B)=P(A)+P(B)-P(A\:and\:B)[/tex]Since our events are mutually exclusive(the student either walks or takes the bus), we have:
[tex]P(A\:and\:B)=0[/tex]Then, our probability is:
[tex]P(A\cup B)=\frac{50}{100}+\frac{10}{100}-0=\frac{60}{100}=\frac{3}{5}[/tex]The answer is:
[tex]P(Take\:the\:bus\cup Walk)=\frac{3}{5}[/tex]Question #3 3) The digits of a 2-digit number differ by 5. If the digits are interchanged and the resulting number is added to the original number, we get 99. Find the original number.
ones number = x
Tens number = y
y>x
Number at the tens place y = (x+5)
original number = 10 (x+5)+x
Interchange digits:= 10x+(x+5)
original number + new number = 99
¨[10(x+5)+x]+ [10x+ (x+5)] =99
Solving for x:
(10x+50+x )+( 10x+x+5) = 99
Combine like terms
(11x+50) + (11x+5) = 99
11x+11x+50+5 =99
22x+55 =99
subtract 55 from both sides
22x +55-55= 99-55
22x = 44
Divide both sides by 22
22x/22= 44/22
x = 2
unit place: 2
tens place = x+5 = 2+5 = 7
original number = 72
Can I have help with this problem? I don't really understand how to graph this
Step 1:
The graph of y = -2 is a horizontal line passing through -2.
Step 2
Ethan's income is 4500 per month a list of some of his expenses appear below what percent of Ethan's expenses is insurance
Ethan's income is given as $4500
she pays $95 in insurance
percentage of Ethan's slary in insurance =
[tex]\begin{gathered} =\frac{95}{4500}\times100 \\ =2.11 \end{gathered}[/tex]Answer=2.11%
if x=10 units, then what is the volume of the cube
Knowing that the solid is a cube, you can use the following formula for calculate its volume:
[tex]V=s^3[/tex]Where "s" is the length of any edge of the cube.
In this case, you can identify that:
[tex]s=x=10units[/tex]Find the slope of the graph of the function at the given point.
Consider the following function:
[tex]f(x)=\text{ }\tan(x)\text{ cot\lparen x\rparen}[/tex]First, let's find the derivative of this function. For this, we will apply the product rule for derivatives:
[tex]\frac{df(x)}{dx}=\tan(x)\cdot\frac{d}{dx}\text{ cot\lparen x\rparen + }\frac{d}{dx}\text{ tan\lparen x\rparen }\cdot\text{ cot\lparen x\rparen}[/tex]this is equivalent to:
[tex]\frac{df(x)}{dx}=\tan(x)\cdot(\text{ - csc}^2\text{\lparen x\rparen})\text{+ \lparen sec}^2(x)\text{\rparen}\cdot\text{ cot\lparen x\rparen}[/tex]or
[tex]\frac{df(x)}{dx}=\text{ -}\tan(x)\cdot\text{ csc}^2\text{\lparen x\rparen+ sec}^2(x)\cdot\text{ cot\lparen x\rparen}[/tex]now, this is equivalent to:
[tex]\frac{df(x)}{dx}=\text{ -2 csc \lparen2x\rparen + 2 csc\lparen2x\rparen = 0}[/tex]thus,
[tex]\frac{df(x)}{dx}=0[/tex]Now, to find the slope of the function f(x) at the point (x,y) = (1,1), lug the x-coordinate of the given point into the derivative (this is the slope of the function at the point):
[tex]\frac{df(1)}{dx}=0[/tex]Notice that this slope matches the slope found on the graph of the function f(x), because horizontal lines have a slope 0:
We can conclude that the correct answer is:
Answer:The slope of the graph f(x) at the point (1,1) is
[tex]0[/tex]For the diagram below, if < 4 = 4x - 2, and < 6 = 2x + 14, what is the value of x?Select one:a.8b.16c.4d.5
x = 8
ExplanationsFrom the line geometry shown, the line a and b are parallel lines while line "t" is the transversal.
Since the horizontal lines are parallel, hence;
[tex]\angle4=\angle6(alternate\text{ exterior angle})[/tex]Given the following parameters
[tex]\begin{gathered} \angle4=4x-2 \\ \angle6=2x+14 \end{gathered}[/tex]Equate both expressions to have:
[tex]\begin{gathered} 4x-2=2x+14 \\ 4x-2x=14+2 \\ 2x=16 \\ x=\frac{16}{2} \\ x=8 \end{gathered}[/tex]Hence the value of x is 8
If the radius of both of the green circles is 10 cm, find the area of the yellow region (outside of the circles but inside the rectangle)
The area of the yellow region if the radius of each of the circles is 10 cm is calculated as: 171.7 cm².
How to Find the Area of Circles and Rectangles?The formula that is used to find the areas of circles and rectangles are given below:
Area of a circle = πr², where r is the radius.Area of a rectangle = length × width.Given the diagram in the attachment which shows the green circles and the rectangle, we can deduce the following:
Radius of the each of the circles (r) = 10 cm
Length of the rectangle = 4(r) = 4(10) = 40 cm
Width of the rectangle = 2(r) = 2(10) = 20 cm
The area of the yellow region = area of the rectangle - area of the 2 circles
= (length × width) - 2(πr²)
Substitute
The area of the yellow region = (40 × 20) - 2(π × 10²)
= 800 - 628.3
= 800 - 628.3
= 171.7 cm²
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are figures A and B congruent? explain your reason
Larry purchased a new combine that cost $260,500, minus a rebate of $5,500, a trade-in of $8,500, and a down payment of $7,000. He takes out a loan for the balance at 8% APR over 4 years. Find the annual payment. (Simplify your answer completely. Round your answer to the nearest cent.)
The annual payment for the loan balance is $72,310.03.
What is the periodic payment?The periodic payment is the amount that is paid per period (yearly, monthly, quarterly, or weekly) to repay a loan or a debt.
The periodic payment can be computed using an online finance calculator, making the following inputs.
N (# of periods) = 4 years
I/Y (Interest per year) = 8%
PV (Present Value) = $239,500 ($260,500 - $5,500 - $8,500 - $7,000)
FV (Future Value) = $0
Results:
PMT = $72,310.03
Sum of all periodic payments = $289,240.13
Total Interest = $49,740.13
Thus, the annual payment that Larry needs to make is $72,310.03.
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Which function below has the following domain and range?Domain: {-9, - 5, 2, 6, 10}Range: { -2, 0, 8}
ANSWER :
A.
EXPLANATION :
From the problem, we have the domain and range :
[tex]\begin{gathered} Domain:\lbrace-9,-5,2,6,10\rbrace \\ Range:\lbrace-2,0,8\rbrace \end{gathered}[/tex]The x coordinates must only have the values of the domain
and the y coordinates must only have the values of the range.
The only option that satisfies this condition is :
[tex]\lbrace(2,0),(-5,-2),(10,8),(6,0),(-9,-2)\rbrace[/tex]I need help with this practice problem Having a tough time solving properly
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
r = 7 sin (2θ)
Step 02:
polar equation:
r = 7 sin (2θ):
r = a sin nθ
n odd ==> n petals
n even ===> 2n petals
n = 2 ===> 2*2 petals = 4 petals
graph:
length of the petals:
r = 7 sin (2θ)
θ = 45°
r = 7 sin (2*45°) = 4.95
The answer is:
4.95
A building is 5 feet tall. the base of the ladder is 8 feet from the building. how tall must a ladder be to reach the top of the building? explain your reasoning.show your work. round to the nearest tenth if necessary.
The ladder must be 9.4 ft to reach the top of the building
Here, we want to get the length of the ladder that will reach the top of the building
Firstly, we need a diagrammatic representation
We have this as;
As we can see, we have a right triangle with the hypotenuse being the length of the ladder
We simply will make use of Pythagoras' theorem which states that the square of the hypotenuse is equal to the sum of the squares of the two other sides
Thus, we have;
[tex]\begin{gathered} x^2=5^2+8^2 \\ x^2=\text{ 25 + 64} \\ x^2\text{ = 89} \\ x=\text{ }\sqrt[]{89} \\ x\text{ = 9.4 ft} \end{gathered}[/tex]