Money = $5000
time = 12 years
investment after 24 years
If the investment doubles every 12 years after 24 years the total amount of money will be $10000.0
Find the coordinates of the circumcenter of triangle PQR with vertices P(-2,5) Q(4,1) and R(-2,-3)
The given triangle has vertices at:
[tex]\begin{gathered} P(-2,5) \\ Q(4,1) \\ R(-2,-3) \end{gathered}[/tex]In the coordinate plane, the triangle looks like this:
There are different forms to find the circumcenter, we are going to use the midpoint formula:
[tex]M(x,y)=(\frac{x1+x2}{2},\frac{y1+y2}{2})[/tex]Apply this formula for each vertice and find the midpoint:
[tex]M_{P,Q}=(\frac{-2+4}{2},\frac{5+1}{2})=(1,3)[/tex]For QR:
[tex]M_{Q,R}=(\frac{4+(-2)}{2},\frac{1+(-3)}{2})=(1,-1)[/tex]For PR:
[tex]M_{P,R}=(\frac{-2+(-2)}{2},\frac{5+(-3)}{2})=(-2,1)[/tex]Now, we need to find the slope for any of the line segments, for example, PQ:
We can apply the slope formula:
[tex]m=\frac{y2-y1}{x2-x1}=\frac{1-5}{4-(-2)}=\frac{-4}{6}=-\frac{2}{3}[/tex]By using the midpoint and the slope of the perpendicular line, find out the equation of the perpendicular bisector line, The slope of the perpendicular line is given by the formula:
[tex]\begin{gathered} m1\cdot m2=-1 \\ m2=-\frac{1}{m1} \\ m2=-\frac{1}{-\frac{2}{3}}=\frac{3}{2}_{} \end{gathered}[/tex]The slope-intercept form of the equation is y=mx+b. Replace the slope of the perpendicular bisector and the coordinates of the midpoint to find b:
[tex]\begin{gathered} 3=\frac{3}{2}\cdot1+b \\ 3-\frac{3}{2}=b \\ b=\frac{3\cdot2-1\cdot3}{2}=\frac{6-3}{2} \\ b=\frac{3}{2} \end{gathered}[/tex]Thus, the equation of the perpendicular bisector of PQ is:
[tex]y=\frac{3}{2}x+\frac{3}{2}[/tex]If we graph this bisector over the triangle we obtain:
Now, let's find the slope of the line segment QR:
[tex]m=\frac{-3-1}{-2-4}=\frac{-4}{-6}=\frac{2}{3}[/tex]The slope of the perpendicular bisector is:
[tex]m2=-\frac{1}{m1}=-\frac{1}{\frac{2}{3}}=-\frac{3}{2}[/tex]Let's find the slope-intercept equation of this bisector:
[tex]\begin{gathered} -1=-\frac{3}{2}\cdot1+b \\ -1+\frac{3}{2}=b \\ b=\frac{-1\cdot2+1\cdot3}{2}=\frac{-2+3}{2} \\ b=\frac{1}{2} \end{gathered}[/tex]Thus, the equation is:
[tex]y=-\frac{3}{2}x+\frac{1}{2}[/tex]This bisector in the graph looks like this:
Now, to find the circumcenter we have to equal both equations, and solve for x:
[tex]\begin{gathered} \frac{3}{2}x+\frac{3}{2}=-\frac{3}{2}x+\frac{1}{2} \\ \text{Add 3/2x to both sides} \\ \frac{3}{2}x+\frac{3}{2}+\frac{3}{2}x=-\frac{3}{2}x+\frac{1}{2}+\frac{3}{2}x \\ \frac{6}{2}x+\frac{3}{2}=\frac{1}{2} \\ \text{Subtract 3/2 from both sides} \\ \frac{6}{2}x+\frac{3}{2}-\frac{3}{2}=\frac{1}{2}-\frac{3}{2} \\ \frac{6}{2}x=-\frac{2}{2} \\ 3x=-1 \\ x=-\frac{1}{3} \end{gathered}[/tex]Now replace x in one of the equations and solve for y:
[tex]\begin{gathered} y=-\frac{3}{2}\cdot(-\frac{1}{3})+\frac{1}{2} \\ y=\frac{1}{2}+\frac{1}{2} \\ y=1 \end{gathered}[/tex]The coordinates of the circumcenter are: (-1/3,1).
In the graph it is:
Use the pair of functions to find f(g(x)) and g(f(x)). Simplify your answers.f(x)=sqrt(x)+2g(x)=x^2+7f(g(x))= ?g(f(x))= ?
Answer:
[tex]\begin{gathered} \begin{equation*} f(g(x))=\sqrt{x^2+7}+2 \end{equation*} \\ \begin{equation*} g(f(x))=x+4\sqrt{x}+11 \end{equation*} \end{gathered}[/tex]Explanation:
Given the functions f(x) and g(x) below:
[tex]\begin{gathered} f(x)=\sqrt{x}+2 \\ g\mleft(x\mright)=x^2+7 \end{gathered}[/tex]Part A
We want to find the simplified form of f(g(x)).
[tex]f(x)=\sqrt{x}+2[/tex]Replace x with g(x):
[tex]f(g(x))=\sqrt{g(x)}+2[/tex]Finally, enter the expression for g(x) and simplify if possible:
[tex]\implies f\mleft(g\mleft(x\mright)\mright)=\sqrt{x^2+7}+2[/tex]Part B
We want to find the simplified form of g(f(x)). To do this, begin with g(x):
[tex]g\mleft(x\mright)=x^2+7[/tex]Replace x with f(x):
[tex]g(f(x))=[f(x)]^2+7[/tex]Finally, enter the expression for f(x) and simplify if possible:
[tex]\begin{gathered} g\mleft(f\mleft(x\mright)\mright)=(\sqrt{x}+2)^2+7 \\ =(\sqrt{x}+2)(\sqrt{x}+2)+7 \\ =x+2\sqrt{x}+2\sqrt{x}+4+7 \\ \implies g(f(x))=x+4\sqrt{x}+11 \end{gathered}[/tex]Therefore:
[tex]\begin{equation*} g(f(x))=x+4\sqrt{x}+11 \end{equation*}[/tex]Find the equation of a line parallel to y=x+6 that passes through the point (8,7)(8,7).
The equation of the line which is parallel to the line y = x + 6, and which passes through the point (8, 7) is; y = x - 1
What are parallel lines in geometry?Parallel lines are lines do not intersect and which while on the same plane, have the same slope.
The given line to which the required line is parallel to is y = x + 6
The point through which the required line passes = (8, 7)
The slope of the given line, y = x + 6, is 1,
The slope of parallel lines are equal, which gives;
The slope of the required line is 1
The equation of the required line in point and slope form is therefore;
y - 7 = 1×(x - 8) = x - 8
y = x - 8 + 7 = x - 1
The equation of the required line in slope–intercept form is; y = x - 1
Learn more about the slope–intercept form of the equation of a straight line here:
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hello can you help me with this math question and this a homework assignment
We know that two vectors are ortogonal if and only if:
[tex]\vec{v}\cdot\vec{w}=0[/tex]where
[tex]\vec{v}\cdot\vec{w}=v_1w_1+v_2w_2[/tex]is the dot product between the vectors.
In this case we have the vectors:
[tex]\begin{gathered} \vec{a}=\langle-4,-3\rangle \\ \vec{b}=\langle-1,k\rangle \end{gathered}[/tex]the dot product between them is:
[tex]\begin{gathered} \vec{a}\cdot\vec{b}=(-4)(-1)+(-3)(k) \\ =4-3k \end{gathered}[/tex]and we want them to be ortogonal, so we equate the dot product to zero and solve the equation for k:
[tex]\begin{gathered} 4-3k=0 \\ 4=3k \\ k=\frac{4}{3} \end{gathered}[/tex]Therefore, for the two vector to be ortogonal k has to be 4/3.
Show instructionsQuestion 1 (1 point)Does the point (0,5) satisfy the equation y = x + 5?TrueFalse
The equation is
[tex]y=x+5[/tex]The point given is:
[tex](x,y)=(0,5)[/tex]The x coordinate given is 0 and the y coordinate given is 5.
We put the respective point and see if the equation holds true or not.
Thus,
[tex]undefined[/tex]In how many ways can Joe, Mary, Steve, Tina and Brenda be seated around a round table?241220
The number of people to be seated around the table, n=5.
Now, n=5 people can be seated in a circle in (n-1)! ways.
[tex](n-1)!=(5-1)=4!\text{ =4}\times3\times2\times1=24[/tex]Therefore, Joe, Mary, Steve, Tina and Brenda can be seated around the round table in 24 ways.
solve for y. 2x-y=12
Answer:
2x - 12 = y
Step-by-step explanation:
→ Add y to both sides
2x = 12 + y
→ Minus 12 from both sides
2x - 12 = y
In the equation y = 2x, y represents the perimeter of a square.What does x represent?Ahalf the length of each sideBthe length of each sideСtwice the length of each sideDtwice the number of sides
Given:
An equation that represents the perimeter of a square:
[tex]y=2x[/tex]To find:
What x represents.
Solution:
It is known that the perimeter of the square is equal to four times the side of the square.
Let the side of the square be s. So,
[tex]\begin{gathered} y=P \\ 2x=4s \\ x=\frac{4s}{2} \\ x=2s \end{gathered}[/tex]Therefore, x represents twice the length of each side.
Can't help me??
x/4 - 9 = 7
solve the equation... use transposing method
The Answer Is x = 64.
Explanation.x/4 - 9 = 7
x/4 = 7 + 9
x/4 = 16
x = 16 × 4
x = 64
_________________
Class: High School
Lesson: Equation
[tex]\boxed{ \colorbox{lightblue}{ \sf{ \color{blue}{ Answer By\:Cyberpresents}}}}[/tex]
Answer:
x = 64
Step-by-step explanation:
x/4 - 9 =7
Step 1: Add 9 to both sides
x/4 - 9 + 9 = 7 + 9
x/4 = 16
Step 2: Multiply right side by 4
x/4= 16 x 4
x = 64
Step 3: Prove your x-value
64/4 = 16 - 9 = 7
64/4 - 9 = 7
So x = 64
PS: Please make brainliest.
Which of the following numbers are not natural numbers?Select one:a. 1,000,000b. 5,032c. 1/4d. 25
Natural numbers are those who you use to count elements, they are by definition positive integers.
C. is not an integer, so it is not a natural number
b. 5032, a. 1000000 and d.25 are positive integers. These are natural numbers.
I need to find out what sine cosine and cotangent is, if this is my reference angle in the picture
We will use the following trigonometric identities
[tex]\begin{gathered} \tan \Theta=\frac{sin\Theta}{\cos \Theta} \\ \cot \Theta=\frac{1}{\tan \Theta} \end{gathered}[/tex]Using these identities we can identify
[tex]\begin{gathered} \tan \Theta=\frac{12}{5} \\ \sin \Theta=12 \\ \cos \Theta=5 \\ \cot \Theta=\frac{1}{\frac{12}{5}}=\frac{5}{12} \end{gathered}[/tex][tex]\begin{gathered} \Theta=\tan ^{-1}(2.4) \\ \Theta=67.38º \\ \sin \Theta=0.92 \\ \cos \Theta=0.38 \end{gathered}[/tex][tex]\begin{gathered} \tan \Theta=\frac{opposite}{\text{adjacent}}^{} \\ \text{opposite}=12 \\ \text{adjacent}=5 \\ \text{hippotenuse=}\sqrt[\square]{12^2+5^2} \\ \text{hippotenuse=}13 \end{gathered}[/tex][tex]\begin{gathered} \sin \Theta=\frac{12}{13} \\ \cos \Theta=\frac{5}{13} \end{gathered}[/tex]Find the equation of the line. Use exortumbers. st V = 2+ 9 8- 6+ 5+ -4 3+ 2+ 1+ T + -9-8-7-65 2 3 5 6 7 8 9 4 -3 -2 -2 + -3+ -4+ -5* -6+ -7+ -8+
We can see that the line passes by the points (0, -5) & (5, 0), using this information we proceed as follows:
1st: We find the slope(m):
[tex]m=\frac{0+5}{5-0}\Rightarrow m=1[/tex]2nd: We use one of the points from the line and the slope to replace in the following expression:
[tex]y-y_1=m(x-x_1)[/tex]That is (Using point (0, -5):
[tex]y+5=1(x-0)[/tex]Now, we solve for y:
[tex]\Rightarrow y=x-5[/tex]And that is the equation of the line shown.
There are 10 males and 18 females in the Data Management class. How many different committees of 5 students can be formed if there must be 3 males and 2 femalesA: 18360B: 2600C: 98280D: 15630
Answer:
A: 18360
Explanation:
The number of ways of combinations to select x people from a group of n people is calculated as
[tex]\text{nCx}=\frac{n!}{x!(n-x)!}[/tex]Since we need to form committees with 3 males and 2 females, we need to select 3 people from the 10 males and 2 people from the 18 females, so
[tex]10C3\times18C2=\frac{10!}{3!(10-3)!}\times\frac{18!}{2!(18-2)!}=120\times153=18360[/tex]Therefore, there are 18360 ways to form a committee.
So, the answer is
A: 18360
What is the slope of a line perpendicular to the line whose equation is15x + 12y = -108. Fully reduce your answer.Answer:Submit Answer
GIven:
The equation of a line is 15x+12y=-108.
The objective is to find the slope of the perpencidular line.
It is known that the equation of straight line is,
[tex]y=mx+c[/tex]Here, m represents the slope of the equation and c represents the y intercept of the equation.
Let's find the slope of the given equation by rearranging the eqation.
[tex]\begin{gathered} 15x+12y=-108 \\ 12y=-108-15x \\ y=-\frac{15x}{12}-\frac{108}{12} \\ y=-\frac{5}{4}x-9 \end{gathered}[/tex]By comparing the obtained equation with equation of striaght line, the value of slope is,
[tex]m_1=-\frac{5}{4}[/tex]THe relationship between slopes of a perpendicular lines is,
[tex]\begin{gathered} m_1\cdot m_2=-1 \\ -\frac{5}{4}\cdot m_2=-1 \\ m_2=-1\cdot(-\frac{4}{5}) \\ m_2=\frac{4}{5}^{} \end{gathered}[/tex]Hence, the value of slope of perpendicular line to the given line is 4/5.
You are given the equation 12 = 3x + 4 with no solution set. Part A: Determine two values that make the equation false. Part B: Explain why your integer solutions are false. Show all work.
[tex]12=3x+4 \\ \\ 8=3x \\ \\ x=8/3[/tex]
So, two integer values are 1 and 2 since they are not the solution to the equation.
3. A student solved an order of operations problem asshown.(2 - 4)2 – 5(6 - 3) + 13(-2)2 - 30 - 3 + 134 - 33 + 13-16What error did this student make? Explain in completesentences. What should the correct answer be?
Applying PEMDAS
P ----> Parentheses first
E -----> Exponents (Powers and Square Roots, etc.)
MD ----> Multiplication and Division (left-to-right)
AS ----> Addition and Subtraction (left-to-right)
Parentheses first
[tex]\begin{gathered} (2-4)=-2 \\ (6-3)=3 \\ \end{gathered}[/tex]substitute
[tex]\begin{gathered} (-2)2-5(3)+13 \\ -4-15+13 \\ -4-2 \\ -6 \\ \end{gathered}[/tex]The student error was misapplication of the comutative property
2. The Venn diagram shows the sets U, X and Y.UXY.34 246..9.512:31List the elements of the following sets:(a) X(b) Y(c) U(d) XUY(e) XnY(g) X\Y(h) Y\X(f) X'(1) (XY)2:31
Given the Venn diagram in the question, we can proceed to answer the questions as follow
[tex]\begin{gathered} X=\text{members of the subset X} \\ This\text{ gives: 1,2,3,4, and 5} \end{gathered}[/tex][tex]\begin{gathered} QuestionA\text{ } \\ X=1,2,3,4,and\text{ 5} \\ \end{gathered}[/tex]Question B
Y= members of subset Y
Y =2,4,6, and 8
Question C
U means that we should list all elements in the universal set
U = ALL members of the set
U = 1,2,3,4,5,6,7,8, and 9
Question D
This is the union of both sets X and Y. This means we will list all the members that are found in the 2 subsets
[tex]\text{XUY}=1,2,3,4,5,6,\text{ and 8}[/tex]Question E
[tex]\begin{gathered} \text{XnY means we are to find the elements that are common to both X and Y} \\ \text{XnY}=2\text{ and 4} \end{gathered}[/tex]Question F
X' means that we should find all members of the set except that of X
[tex]X^{\prime}=6,7,8,\text{ and 9}[/tex]Question G
X\Y means that we should list the elements of X that are not found in Y
X\Y= 1,3, and 5
Question H
Y\X means that we should list the elements of Y that are not found in X
Y\X= 6, and 7
Question I
To solve (XnY)' we will follow the steps below
Step 1: Find (XnY)
[tex]\text{XnY}=2\text{ and 4}[/tex]Step 2: Find (XnY)'
[tex]We\text{ will list all elements aside (XnY)}[/tex][tex](XnY)^{^{\prime}}\Rightarrow1,3,5,6,7,8,\text{and 9}[/tex]
• 5th 3230 [] What would be the slope of a line perpendicular to the line graphed above? -2 2 1/2 -1/2 Zoom in Double Jeop 3:39
When two lines are perpendicular to each other, their slopes would be a negative inverse of each other. This simply means the slope of a line perpendicular to the one in the question should be equal to the negative inverse of the one we have here. Let us begin by calculating the slope of this line.
When you are given two endpoints, the slope is derived as;
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]When the coordinates are (0, 3) and (-1.5, 0)
That is, when x = 0 then y = 3 (observe the point where the line touches the vertical axis), and when x = -1.5, then y = 0 (observe the point where the line touches the horizontal axis)
Therefore, the coordinates (0, 3) and (-1.5, 0) are now your (x1, y1) and (x2, y2)
[tex]\begin{gathered} m=\frac{0-3}{-1.5-0} \\ m=\frac{-3}{-1.5} \\ m=2 \end{gathered}[/tex]Therefore the slope of a line perpendicular to the one on the graph is -1/2.
A student council president wants to learn about the preferred theme for the upcoming spring dance. Select all the samples that are representative of the population.
The idea of being representative of the population is actually reflecting the characteristics (features) we want to study of the whole population.
In this case, the samples that better represent the whole population are B and D. Why? Because they give us the possibility of taking a student of every grade. The other options, excluding the "bus option" and the first option, fail doing that. Finally, these options (bus option and lunch option) are related to the council president.
Find the equation of the line connecting the points (2,0) and (3,15). Write your final answer in slope-intercept form.
The first step to find the equation of the line is to find its slope. To do it, we need to use the following formula:
[tex]m=\frac{y2-y1}{x2-x1}[/tex]Where y2 and y1 are the y coordinates of 2 given points on the line, and x2 and x1 are the x coordinates of the same points. m is the slope.
Replace for the given values and find the slope:
[tex]m=\frac{15-0}{3-2}=\frac{15}{1}=15[/tex]Now, use one of the given points and the slope in the point slope formula:
[tex]y-y1=m(x-x1)[/tex]Replace for the known values:
[tex]\begin{gathered} y-0=15(x-2) \\ y=15x-30 \end{gathered}[/tex]The equation of the line is y=15x-30
Does the point (3,-1) lie on the circle (x + 1)2 + (y - 1)1)2 = 16?no; the point is not represented by (h, k) in the equationyes; when you plug the point in for x and y you get a true statementno; when you plug in the point for x and y in the equation, you do not get a trueyes; the point is represented by (h, k) in the equation
We are given an equation of a circle and a point. We are then asked to find if the point lies on the circle. The equation of the circle and the point is given below
[tex]\begin{gathered} \text{Equation of the circle} \\ (x+1)^2+(y-1)^2=16 \\ \text{Given point =(3,-1)} \end{gathered}[/tex]To find if the point lies in the circle, we can use the simple method of substituting the coordinates into the equation of the circle.
This can be seen below:
[tex]\begin{gathered} (3+1)^2+(-1-1)^2=16 \\ 4^2+(-2)^2=16 \\ 16+4=16 \\ \therefore20\ne16 \end{gathered}[/tex]Since 20 cannot be equal to 16, this implies that the point does not lie on the circle.
ANSWER: Option 3
Each of 6 students reported the number of movies they saw in the past year. This is what they reported:20, 16, 9, 9, 11, 20Find the mean and median number of movies that the students saw.If necessary, round your answers to the nearest tenth.
Solution:
The number of movies 6 students reported they saw in the past year is;
[tex]20,16,9,9,11,20[/tex]The mean number of movies that the students saw is;
[tex]\begin{gathered} \bar{x}=\frac{\Sigma x}{n} \\ n=6 \\ \bar{x}=\frac{20+16+9+9+11+20}{6} \\ \bar{x}=\frac{85}{6} \\ \bar{x}=14.2 \end{gathered}[/tex]The mean round to the nearest tenth is 14.2
Also, the median is the middle number of the data set after arranging either in ascending or descending order.
[tex]9,9,11,16,20,20[/tex]The median number of movies the students saw is the average or mean of the third and fourth.
[tex]\begin{gathered} \text{Median}=\frac{11+16}{2} \\ \text{Median}=\frac{27}{2} \\ \text{Median}=13.5 \end{gathered}[/tex]The median round to the nearest tenth is 13.5
I really need help solving this practice from my prep guide in trigonometry
Given: Different angles in degrees and in terms of pi. The different angles are:
[tex]\begin{gathered} a)714^0 \\ b)\frac{23\pi}{5} \\ c)120^0 \\ d)\frac{31\pi}{6} \end{gathered}[/tex]To Determine: The equivalence of the given angles
The equivalent of degree and pi is given as
[tex]\begin{gathered} 2\pi=360^0 \\ \pi=\frac{360^0}{2} \\ \pi=180^0 \\ 360^0=2\pi \\ 1^0=\frac{2\pi}{360^0} \\ 1^0=\frac{1}{180}\pi \end{gathered}[/tex][tex]\begin{gathered} a)714^0 \\ 1^0=\frac{1}{180}\pi \\ 714^0=\frac{714^0}{180^0}\pi \\ 714^0=3\frac{29}{30}\pi \\ 714^0=\frac{119\pi^{}}{30} \end{gathered}[/tex][tex]\begin{gathered} b)\frac{23\pi}{5} \\ 1\pi=180^0 \\ \frac{23\pi}{5}=\frac{23}{5}\times180^0 \\ \frac{23\pi}{5}=828^0 \end{gathered}[/tex][tex]\begin{gathered} c)120^0 \\ 1^0=\frac{\pi}{180} \\ 120^0=120\times\frac{\pi}{180} \\ 120^0=\frac{2\pi}{3} \end{gathered}[/tex][tex]\begin{gathered} d)\frac{31\pi}{6} \\ 1\pi=180^0 \\ \frac{31\pi}{6}=\frac{31}{6}\times180^0 \\ \frac{31\pi}{6}=930^0 \end{gathered}[/tex]ALTERNATIVELY
A revolution is 360 degree
[tex]\begin{gathered} a)714^0 \\ \text{Multiples of 360 degre}e \\ 2\times360^0=720^0 \\ \text{equivalent of 714 degre}e\text{ would be} \\ 720^0-714^0=6^0 \end{gathered}[/tex][tex]undefined[/tex][tex]\begin{gathered} a)714^0=\frac{119\pi}{30} \\ b)\frac{23\pi}{5}=828^0 \\ c)120^0=\frac{2\pi}{3} \\ d)\frac{31\pi}{6}=930^0 \end{gathered}[/tex]reduce the square root of -360
reduce the square root of
[tex]\begin{gathered} \sqrt[]{-360} \\ 360=36\cdot10=6^2\cdot10 \\ \end{gathered}[/tex]There is no square root for the negative number
so, this is represent a complex number
So,
[tex]\begin{gathered} \sqrt[]{-360}=\sqrt[]{-1}\cdot\sqrt[]{360} \\ =i\cdot\sqrt[]{6^2\cdot10} \\ =i\cdot6\sqrt[]{10} \\ =6\sqrt[]{10}\cdot i \end{gathered}[/tex]Write the equation of a line, in slope-intercept form, that has a slope of m= -2 and y-interceptof b = -8.Y=
Explanation
We are given the following:
[tex]\begin{gathered} slope(m)=-2 \\ y\text{ }intercept(b)=-8 \end{gathered}[/tex]We are required to determine the equation of the line in the slope-intercept form.
We know that the equation of a line in slope-intercept form is given as:
[tex]\begin{gathered} y=mx+b \\ where \\ m=slope \\ b=y\text{ }intercept \end{gathered}[/tex]Therefore, we have:
[tex]\begin{gathered} y=mx+b \\ where \\ m=-2\text{ }and\text{ }b=-8 \\ y=-2x+(-8) \\ y=-2x-8 \end{gathered}[/tex]Hence, the answer is:
[tex]y=-2x-8[/tex]A woman is floating in a
boat that is 175 feet from
the base of a cliff. The cliff
is 250 feet tall. What is the
angle of elevation from
the boat to the top of the
cliff?
The angle of depression between the cliff and the boat is 55.0
What is angle of depression?
The angle of depression is the angle between the horizontal line and the observation of the object of from the horizontal line. It's basically used to get the of distance of the two objects where the angles and an of object's distance from the ground are known to us.
A boat is moving 175 feet from the base and a women is in the boat.the height of the cliff is 259 feet tall. Here we have to find the angle between the cliff and the boat.
As per the given question
We have a right angled traingle where base is 175 ft and height is 250 ft.
Thus,
We know that tan theta =opposite/adjacent
250/175
So theta=tan^-1(250/175)
So theta = 55.0
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A rectangle has a diagonal of length 10 cm and a base of length 8 cm . Find its height
Given:
The length of diagonal of rectangle is d = 10 cm.
The length of base is b = 8 cm.
Explanation:
The relation between length, height and diagonal of rectangle is given by pythagoras theorem. So
[tex]d^2=l^2+h^2[/tex]Substitute the values in the equation to obtain the value of h.
[tex]\begin{gathered} (10)^2=(8)^2+h^2 \\ 100=64+h^2 \\ h=\sqrt[]{100-64} \\ =\sqrt[]{36} \\ =6 \end{gathered}[/tex]So the height of rectangle is 6 cm.
Answer: 6 cm
What is (are) the solution(s) to the system of equations y = -x + 4 and y = -x^2 + 4 ?
Given:
[tex]\begin{gathered} y=-x+4----(1) \\ y=-x^2+4----(2) \end{gathered}[/tex]Required:
To find the solutions to the given equations.
Explanation:
Put equation 1 in 2, we get
[tex]\begin{gathered} -x+4=-x^2+4 \\ \\ -x+4+x^2-4=0 \\ \\ x^2-x=0 \\ \\ x(x-1)=0 \\ \\ x=0,1 \end{gathered}[/tex]When x=0,
[tex]\begin{gathered} y=-0+4 \\ y=4 \end{gathered}[/tex]When x=1,
[tex]\begin{gathered} y=-1+4 \\ =3 \end{gathered}[/tex]Final Answer:
The solution are
[tex]x=0,1[/tex]The solution sets are
[tex]\begin{gathered} (0,4)\text{ and} \\ (1,3) \end{gathered}[/tex]Which of the following describes point D?
Answer:
(0,4)
Step-by-step explanation:
Hi! :)
I am Pretty sure this is what it is, if this is not what you are needing please let me know.
Can you help me with this true and false problem?
FALSE.
Explanations:Given the linear relations 2x - 3y = 4 and y = -2/3 x + 5
Both equations are equations of a line. For the lines to be perpendicular, the product of their slope is -1
The standard equation of a line in slope-intercept form is expressed as
[tex]y=mx+b[/tex]m is the slope of the line
For the line 2x - 3y = 4, rewrite in standard form
[tex]\begin{gathered} 2x-3y=4 \\ -3y=-2x+4 \\ y=\frac{-2}{-3}x-\frac{4}{3} \\ y=\frac{2}{3}x-\frac{4}{3} \end{gathered}[/tex]Compare with the general equation
[tex]\begin{gathered} mx=\frac{2}{3}x \\ m=\frac{2}{3} \end{gathered}[/tex]The slope of the line 2x - 3y = 4 is 2/3
For the line y = -2/3 x + 5
[tex]\begin{gathered} mx=-\frac{2}{3}x \\ m=-\frac{2}{3} \end{gathered}[/tex]The slope of the line y = -2/3 x + 5 is -2/3
Take the product of their slope to determine whether they are perpendicular
[tex]\begin{gathered} \text{Product = }\frac{2}{3}\times-\frac{2}{3} \\ \text{Product = -}\frac{4}{9} \end{gathered}[/tex]Since the product of their slope is not -1, hence the linear relations do not represent lines that are perpendicular. Hence the correct answer is FALSE