Answer:
∅ = 44°
Step-by-step explanation:
cos∅ = sin46°
∅ = (90 - 46)°
∅ = 44°
Hope this helps
A spinner is divided into 5 equally sized segments colored blue, green, black, red, and yellow. Suppose you spin the wheel once and then spin it again. What is the probability of landing on the color red both times? Give your answer as an exact fraction and reduce the fraction as much as possible.
The probability of landing in any colour is:
[tex]\frac{1}{5}[/tex]so if we want to get a red, both times, we to multiply 1/5 twice
[tex]\frac{1}{5}\cdot\frac{1}{5}=\frac{1}{25}[/tex]so, the probability of getting red twice is 1/25
In the figure, ∆ABD ≅ ∆CBD by Angle-Side-Angle (ASA). Which segments are congruent by CPCTC? BC ≅ AD CB ≅ AB AB ≅ CD DB ≅ DC
By CPCTC this is the only valid answer:
CB ≅ AB
Another statement should be AD≅ CD
Solve the system of equation by the elimination method {1/3x+1/2y=1/2{1/6x-1/3y=5/6(x,y)=(_, _)
Solution
- The solution steps to solve the system of equations by elimination is given below:
[tex]\begin{gathered} \frac{x}{3}+\frac{y}{2}=\frac{1}{2}\text{ \lparen Equation 1\rparen} \\ \\ \frac{x}{6}-\frac{y}{3}=\frac{5}{6}\text{ \lparen Equation 2\rparen} \\ \\ \text{ Multiply Equation 2 by 2} \\ 2\times(\frac{x}{6}-\frac{y}{3})=\frac{5}{6}\times2 \\ \\ \frac{x}{3}-\frac{2y}{3}=\frac{5}{3}\text{ \lparen Equation 3\rparen} \\ \\ \\ \text{ Now, }\frac{x}{3}\text{ is common to both Equations 1 and 3.} \\ \\ \text{ We can therefore subtract both equations to eliminate }x. \\ \text{ We have:} \\ \text{ Equation 1 }-\text{ Equation 3} \\ \\ \frac{x}{3}+\frac{y}{2}-(\frac{x}{3}-\frac{2y}{3})=\frac{1}{2}-\frac{5}{3} \\ \\ \frac{x}{3}-\frac{x}{3}+\frac{y}{2}+\frac{2y}{3}=\frac{1}{2}-\frac{5}{3}=\frac{3}{6}-\frac{10}{6} \\ \\ \frac{y}{2}+\frac{2y}{3}=-\frac{7}{6} \\ \\ \frac{3y}{6}+\frac{4y}{6}=-\frac{7}{6} \\ \\ \frac{7y}{6}=-\frac{7}{6} \\ \\ \therefore y=-1 \\ \\ \text{ Substitute the value of }y\text{ into any of the equations, we have:} \\ \frac{1}{3}x+\frac{1}{2}y=\frac{1}{2} \\ \frac{1}{3}x+\frac{1}{2}(-1)=\frac{1}{2} \\ \\ \frac{1}{3}x=\frac{1}{2}+\frac{1}{2} \\ \\ \frac{1}{3}x=1 \\ \\ \therefore x=3 \end{gathered}[/tex]Final Answer
The answer is:
[tex]\begin{gathered} x=3,y=-1 \\ \\ \therefore(x,y)=(3,-1) \end{gathered}[/tex]12. Jimmy is paid $14.50 per hour for a regular forty-hour work week and 5 point time and a half for any hour worked over 40. This pas week, Jimmy earned $754.00 in total pay. How many hours of overtime did Jimmy work?
Jimmy is paid $14.50 per hour for a regular forty-hour work week and 5 point time and a half for any hour worked over 40. This pas week, Jimmy earned $754.00 in total pay. How many hours of overtime did Jimmy work?
Let
x -----> the total hours worked
we have that
$14.50 --------> 40 hours
5.5($14.50) -------> > 40 hours
so
754=14.50*40+5.5(14.50)x
solve for x
754=580+79.75x
79.75x=754-580
79.75x=174
x=2.2 hours
Copper has a density of 4.44 g/cm3. What is the volume of 2.78 g of copper?
60 points please help
The volume of 2.78 g of copper is 0.626 [tex]cm^{3}[/tex].
According to the question,
We have the following information:
Density of cooper = 4.44 [tex]g/cm^{3}[/tex]
Mass of copper = 2.78 g
We know that the following formula is used to find the density of any material:
Density = Mass/volume
Let's denote the volume of copper be V.
Now, putting the values of mass and density here:
4.44 = 2.78/V
V = 2.78/4.44
V = 0.626 [tex]cm^{3}[/tex]
(Note that the units if mass, volume and density are written with the numbers. For example, in this case, the unit of mass is grams, the unit of volume is [tex]cm^{3}[/tex].)
Hence, the volume of the copper is 0.626 [tex]cm^{3}[/tex].
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What is the solution of the system of equations? Explain.18x+15-y=05y=90x+12
The given system is
[tex]\begin{cases}18x+15-y=0 \\ 5y=90x+12\end{cases}[/tex]First, we multiply the first equation by 5.
[tex]\begin{cases}90x+75-5y=0 \\ 5y=90x+12\end{cases}[/tex]Then, we combine the equations
[tex]\begin{gathered} 90x+75-5y+5y=90x+12 \\ 90x+75=90x+12 \\ 75=12 \end{gathered}[/tex]Given that the result is not true (75 is not equal to 12), we can deduct that the system has no solutions.
In the accompanying regular pentagonal prism, suppose that each base edge measures 7 in. and that the apothem of the base measures 4.8 in. The altitude of the prism measures 10 in.A regular pentagonal prism and a pentagon are shown side by side. The pentagon contains a labeled segment and angle.The prism contains a horizontal top and bottom and vertical sides. The front left face and front right face meet the bottom base at right angles.The pentagon is labeled "Base".A line segment starts in the center of the pentagon, travels down vertically, and ends at the edge. The segment is labeled a.The vertical segment forms a right angle with the edge.(a)Find the lateral area (in square inches) of the prism.in2(b)Find the total area (in square inches) of the prism.in2(c)Find the volume (in cubic inches) of the prism.in3
To determine the lateral area of the prism;
[tex]Lateral\text{ area=perimeter of the base}\times height[/tex][tex]Lateral\text{ area=5\lparen7\rparen }\times10=350in^2[/tex]To determine total area of the prism;
[tex]Total\text{ area=2\lparen area of base\rparen+Lateral area}[/tex][tex]\begin{gathered} Total\text{ area of the prism=2\lparen}\frac{1}{2}\times perimeter\text{ of the base}\times apotherm\text{\rparen+350} \\ \end{gathered}[/tex][tex]\begin{gathered} Total\text{ area of the prism=2\lparen}\frac{1}{2}\times5(7)\times4.8\text{\rparen+380=168+350=518in}^2 \\ \end{gathered}[/tex]To determine the volume of the prism;
[tex]Volume\text{ = base area }\times height[/tex][tex]Volume=\frac{1}{2}\times5(7)\times4.8\times10=840in^3[/tex]Hence
Here's a graph of a linear function. Write theequation that describes that function.Express it in slope-intercept form.Enter the correct answer.000DONEClear allĐOO
A train leaves a station and travels north at a speed of 175 km/h. Two hours later, a second train leaves on a parallel track and travels north at 225 km/h. How far from the station will theymeet?The trains will meet (?) away from the station(Type an integer or a decimal.)
Given,
The speed of first train is 175km/h.
The speed of second train is 225 km/h.
Consider, the time taken by first train to meet is x h.
The time taken by the second train to meet is (x-2) h.
The distance is calculated as,
[tex]\text{Distance}=\text{speed}\times time[/tex]At the meeting point the distance covered by both train is same,
So, taking the distance equation of both trains in equal,
[tex]\begin{gathered} 175\times x=225\times(x-2) \\ 175x=225x-450 \\ -50x=-450 \\ x=9 \end{gathered}[/tex]The first train meet to second train after distance,
[tex]d=175\times9=1575\text{km}[/tex]The second train meet to first train after distance,
[tex]d=225\times(9-2)=1575\text{ km}[/tex]Hence, both train meet after 1575 km from the station.
Inga is solving 2x2 + 12x – 3 = 0. Which steps could she use to solve the quadratic equation? Select three options. 2(x2 + 6x + 9) = 3 + 18 2(x2 + 6x) = –3 2(x2 + 6x) = 3 x + 3 = Plus or minus StartRoot StartFraction 21 Over 2 EndFraction EndRoot 2(x2 + 6x + 9) = –3 + 9
Inga should use the first option that is [tex]2(x^2+6x+9)=3+18[/tex] to solve the quadratic equation.
Given equation:-
[tex]2x^2+12x-3=0[/tex]
We have to find which one of the given options are needed to solve the quadratic equation.
The given quadratic equation can be rewritten as:-
[tex]2x^2+12x[/tex]-3+3=0+3
[tex]\\2x^2+12x[/tex]-3 + 3 + 18=0 +3 +18
[tex]2x^2[/tex] + 12x + 18 = 0 + 3 + 18
Hence, the answer is the first option.
Quadratic equation
Quadratic equations are the polynomial equations of degree 2 in one variable of type [tex]f(x) = ax^2 + bx + c = 0[/tex]where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where ‘a’ is called the leading coefficient and ‘c’ is called the absolute term of f (x). The values of x satisfying the quadratic equation are the roots of the quadratic equation (α, β).
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3. Ketin's card collection is made up of baseball cards and footbal cards. The ratio of baseball cards to football cards is 6 to 7. He has 120 baseball cards. How many cards are in Kerin's card collection? Show your work.
SOLUTION
Let the total number of cards in Ketin's card collection be k
Let the number of baseball cards be b, and
the number of football cards be f
Now, the ratio of baseball cards to football cards is 6 to 7, that is
[tex]\begin{gathered} b\colon f=6\colon7 \\ \frac{b}{f}=\frac{6}{7} \\ \text{cross multiplying, we have } \\ 7\times b=6\times f \\ 7b=6f \\ \text{dividing both sides by 7 to get b, we have } \\ \frac{7b}{7}=\frac{6f}{7} \\ b=\frac{6f}{7} \end{gathered}[/tex]Also, he has 120 baseball cards.
This means
[tex]\begin{gathered} b=120 \\ \text{but } \\ b=\frac{6f}{7} \\ \text{That means that } \\ b=\frac{6f}{7}=120 \\ So,\text{ } \\ \frac{6f}{7}=120 \\ \frac{6f}{7}=\frac{120}{1} \\ \text{cross multiplying, we have } \\ 6f=120\times7 \\ \text{dividing by 6, we have } \\ f=\frac{120\times7}{6} \\ 120\text{ divided by 6 = 20, we have } \\ f=20\times7 \\ f=140 \end{gathered}[/tex]So, the total number of cards in Ketin's card collection is
[tex]120+140=260[/tex]Hence the answer is 260
How do I solve:7/8+y= -1/8
We are given the following equation
[tex]\frac{7}{8}+y=-\frac{1}{8}[/tex]Let us solve the equation for variable y
Our goal is to separate out the variable y
Subtract 7/8 from both sides of the equation.
[tex]\begin{gathered} \frac{7}{8}-\frac{7}{8}+y=-\frac{1}{8}-\frac{7}{8} \\ y=-\frac{1}{8}-\frac{7}{8} \end{gathered}[/tex]Since the denominators of the two fractions are the same, simply add the numerators.
[tex]\begin{gathered} y=-\frac{1}{8}-\frac{7}{8} \\ y=\frac{-1-7}{8} \\ y=\frac{-8}{8} \\ y=-1 \end{gathered}[/tex]Therefore, the value of y is -1
Instructions: Find the area of the circle. Round your answer to the nearest tenth.
Given:
The Radius of the circle: 2.5 inch
To find:
The area of the circle
Step-by-step solution:
We know that:
The Area of the circle = π(r)²
The Area of the circle = π(2.5)²
The Area of the circle = 3.14 × (2.5)²
The Area of the circle =
The cost of a pair of skis to a store owner was $700, and she sold the pair of skis for $1020.Step 3 of 3: What was her percent of profit based on selling price? Follow the problem-solving process and round your answer to thenearest hundredth if necessary.
Answer:
Explanation:
• The ,cost price ,of the pair of skis = $700
,• The ,selling price ,of the pair of skis = $1020
To calculate the percentage of profit, use the formula below:
[tex]\text{Percent of Profit=}\frac{Selling\text{ Price-Cost Price}}{\text{Selling Price}}\times\frac{100}{1}[/tex]Substitute the given values:
[tex]\text{Percent of Profit=}\frac{1020\text{-7}00}{\text{7}00}\times\frac{100}{1}\text{=}\frac{320}{\text{7}00}\times\frac{100}{1}=45.71\%[/tex]The percentage profit is % (correct to the nearest hundredth).
Write a numerical expression for the word expression.The product of 2 groups of 7
The given statement is the product of 2 groups of 7.
This statement can be expressed as
[tex]7\cdot7[/tex]Where each seven represent one group.
Find the solution 5(x-9)+3=5x-42A) x=9B) x=-9C) Infinite SolutionsD) No Solutions
Answer:
C. Infinite Solutions
Explanation:
Given the equation
[tex]5\mleft(x-9\mright)+3=5x-42[/tex]First, open the bracket
[tex]\begin{gathered} 5x-45+3=5x-42 \\ 5x-42=5x-42 \end{gathered}[/tex]Since the left-hand side equals the right-hand side, the system has Infinite Solutions.
Find the measures of the sine and cosine of the following triangles
Let x be the side opposite to angle 62 degrees
Let y be the adjacent angle.
The sine of the angle is given as follows:
[tex]\begin{gathered} \sin62=\frac{Opposite}{Hypotenuse}=\frac{x}{10} \\ \end{gathered}[/tex]The cosine is given as:
[tex]\cos62=\frac{Adjacent}{Hypotenuse}=\frac{y}{10}[/tex]A true-false test contains 10 questions. In how many different ways can this test be completed?(Assume we don't care about our scores.)This test can be completed indifferent ways.
Explanation:
The test contains 10 questions, each one can be answered either with 'true' or with 'false' which means that for each question there are only 2 options.
We need to find the number of ways in which the test can be completed.
To answer the question we use the fundamental counting principle:
In this case, there are 2 ways to complete each question, therefore, we multiply that by the 10 questions that we have:
[tex]2\times2\times2\times2\times2\times2\times2\times2\times2\times2[/tex]This can be simplified to
[tex]2^{10}[/tex]which is equal to:
[tex]2^{10}=\boxed{1024}[/tex]Answer:
1024
Tim and Kevin each sold candies and peanuts for a school fund-raiser. Tim sold 16 boxes of candies and 4 boxes of peanuts and earned $132. Kevin sold 6 boxes of peanuts and 20 boxes of candies and earned $190. Find the cost of each. Cost of a box of candy. Cost of a box of peanuts.
We have the following:
let x cost of a box of candy
let y cost of a box of peanuts
[tex]\begin{gathered} \text{ Tim} \\ 16x+4y=132 \\ \text{ Kevin} \\ 20x+6y=190 \end{gathered}[/tex]resolving the system of equations:
[tex]\begin{gathered} 20x+6y=190 \\ 16x+4y=132\Rightarrow4y=132-16x\Rightarrow y=\frac{132-16x}{4} \\ \text{replacing:} \\ 20x+6\cdot(\frac{132-16x}{4})=190 \\ 20x+198-24x=190 \\ -4x=190-198 \\ x=\frac{-8}{-4} \\ x=2 \end{gathered}[/tex]now, for y
[tex]\begin{gathered} y=\frac{132\cdot16\cdot2}{4} \\ y=25 \end{gathered}[/tex]Therefore the cost of the box of candy is $ 2 and the cost of the box of peanuts is $ 25
select the ordered pair that represent solutions of the system of inequalities A) (-1,3)B) (0,0)C) (-4,1)D) (0,2)E) (-5,0)F) ( -4,4)G) (-5,3)H) (-6,5)please answer fast
From the graph, the solution of the graph is given by the area of intersection of the two functions. The are under the area of the two functions are (-4, 1), (-5, 0), (-4, 4)
Haley spent 1/2 oven hour playing on her phone that used up 1/9 of her battery how long would she have to play on her phone to use the entire battery
1/2 hour playing -- 1/9 battery
1 hour playing -- 2/9 battery
1 1/2 hours playing --- 3/9 battery
2 hours playing ------ 4/9 battery
2 1/2 hours playing ---- 5/9 battery
3 hours playing ----- 6/9 battery
3 1/2 hours playing --- 7/9 battery
4 hours playing -----8/9 battery
4 1/2 hours playing ---- 9/9 battery
9/9 represent the entire battery so che can play 4.5 hours on her phone
it can be represented into a fraction as
[tex]4.5=4\frac{1}{2}=\frac{9}{2}[/tex]You have a pizza with a diameter of 6 1/3 in., and a square box that is 6.38 in. Is the box big enough to fit the pizza inside?
Pizza diameter D is given in mixed form:
[tex]\begin{gathered} D=6\frac{1}{3}\text{ in} \\ in\text{ fraction form:} \\ D=\frac{19}{3}\text{ in} \\ In\text{ decimals, } \\ D=6.33\text{ in} \end{gathered}[/tex]Now, me must compare D with the lenght of the square box.
Since the lenght of the box is L=6.38 in. Hence, the box is big enough to
fit the pizza.
[tex]\begin{gathered} \\ \\ \\ \end{gathered}[/tex]What is the row echelon form of this matrix?
The row echelon form of the matrix is presented as follows;
[tex]\begin{bmatrix}1 &-2 &-5 \\ 0& 1 & -7\\ 0&0 &1 \\\end{bmatrix}[/tex]
What is the row echelon form of a matrix?The row echelon form of a matrix has the rows consisting entirely of zeros at the bottom, and the first entry of a row that is not entirely zero is a one.
The given matrix is presented as follows;
[tex]\begin{bmatrix}-3 &6 &15 \\ 2& -6 & 4\\ 1&0 &-1 \\\end{bmatrix}[/tex]
The conditions of a matrix in the row echelon form are as follows;
There are row having nonzero entries above the zero rows.The first nonzero entry in a nonzero row is a one.The location of the leading one in a nonzero row is to the left of the leading one in the next lower rows.Dividing Row 1 by -3 gives:
[tex]\begin{bmatrix}1 &-2 &-5 \\ 2& -6 & 4\\ 1&0 &-1 \\\end{bmatrix}[/tex]
Multiplying Row 1 by 2 and subtracting the result from Row 2 gives;
[tex]\begin{bmatrix}1 &-2 &-5 \\ 0& -2 & 14\\ 1&0 &-1 \\\end{bmatrix}[/tex]
Subtracting Row 1 from Row 3 gives;
[tex]\begin{bmatrix}1 &-2 &-5 \\ 0& -2 & 14\\ 0&2 &4 \\\end{bmatrix}[/tex]
Adding Row 2 to Row 3 gives;
[tex]\begin{bmatrix}1 &-2 &-5 \\ 0& -2 & 14\\ 0&0 &18 \\\end{bmatrix}[/tex]
Dividing Row 2 by -2, and Row 3 by 18 gives;
[tex]\begin{bmatrix}1 &-2 &-5 \\ 0& 1 & -7\\ 0&0 &1 \\\end{bmatrix}[/tex]
The above matrix is in the row echelon form
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help meeeee pleaseeeee!!!
thank you
Step-by-step explanation:
what is the problem ?
first you need to put "1" in place of the x and calculate, and then you need to put "2" in place of the x and calculate.
just simple calculation !
(a)
R(1) = 1000×1² / (1² + 4) = 1000 / 5 = 200
$200 million
(b)
R(2) = 1000×2² / (2² + 4) = 4000 / 8 = 500
$500 million
there ! that's all that was needed.
e) A client takes 1 1/2 tablets of medication three times per day for 4 days. How many tablets will the clienthave taken at the end of four days? Explain how your arrived at your answer.
Given
Client takes 1 1/2 of medication one time
Find
how many tablets will client have been taken at the end of 4 days.
Explanation
Client takes medication one time = 1 1/2
Client takes medication three times =
[tex]\begin{gathered} 1\frac{1}{2}\times3 \\ \frac{3}{2}\times3 \\ \frac{9}{2} \end{gathered}[/tex]medication for 4 days =
[tex]\begin{gathered} \frac{9}{2}\times4 \\ 18 \end{gathered}[/tex]Final Answer
The client takes 18 tablets at the end of four days.
help meeeee pleaseeeee!!!
thank you
Answer:
(f o g) = 464
Step-by-step explanation:
f(x) = x² - 3x + 4; g(x) = -5x
(f o g)(4) = f(g(4))
f(g(4)) = -5(4) = -20
f(g(4)) = (-20)² - 3(-20) + 4
f(g(4)) = 464
I hope this helps!
A glacier in Republica was observed to advance 22inches in a 15 minute period. At that rate, how many feet will the glacier advance in one year?
To fins the rate in feet/year we must change first the measurements to the units required
inches to feat
minutes to years
[tex]22in\cdot\frac{1ft}{12in}=\frac{11}{6}ft[/tex][tex]15\min \cdot\frac{1h}{60\min}\cdot\frac{1day}{24h}\cdot\frac{1year}{365\text{days}}=\frac{1}{35040}\text{years}[/tex]to find the rate divide the distance over the time
[tex]\frac{\frac{11}{6}ft}{\frac{1}{35040}\text{year}}=\frac{11\cdot35040ft}{6\text{year}}=\frac{385440}{6}=\frac{64240ft}{\text{year}}[/tex]IIIALGEBRA AND GEOMETRY REVIEWUsing distribution and combining like teSimplify.-3(x + 1)- 5
-3(x + 1) - 5
(-3)(x) + (-3)(1) -5
-3x - 3 -5
-3x - 8
-2(u + 3) + 4u
-2u - 6 + 4u
2u - 6
The first year shown the number of students per teacher fell below 16 was
Using the y axis, we want to find when it goes below 16
The x value when y is less than 16 for the first time is 2002
The ratio of a quarterback's completed passes to attempted passes is 5 to 8. If he attempted 16 passes, find how many passes he completed. Round to the nearest whole number if necessary.
The ratio of a quarterback's completed passes to attempted passes is 5 to 8. If he attempted 16 passes, find how many passes he completed. Round to the nearest whole number if necessary.
Let
x -----> number of quarterback's completed passes
y -----> number of quarterback's attempted passes
so
x/y=5/8 -----> x=(5/8)*y -----> equation A
y=16 -----> equation B
substitute equation B in equation A
x=(5/8)*16
x=10
therefore
the answer is 10 completed passes