All categories

3 years ago

Brainliest for correct answer

Mathematics

1 answer:

Vladimir79 [104]3 years ago

8 0

Answer:

Step-by-step explanation:

The pre-image is the starting coordinates and it then gets larger. when you multiply four by 3 you get 12 and when you multiply four by -4 you get 16. So the pre-image dilated by 4.

You might be interested in

1/(cot^2 (x)) - 1/(cos^2 (x))

Airida [17]

Step-by-step explanation:

1 is your answer its law of trig identises

5 0

3 years ago

Read 2 more answers

Write a mathematical sentence that expresses the information given below. Use t as your variable name. If necessary:Susanne chec

Feliz [49]

Answer:

$t+8>82^{\circ}$

Explanation:

We were given that:

Let "temperature" be represented by "t"

Suzan checks the temperature at 11:00. If the temperature rises by 8 more degrees, it will break the record high temperature for the day which is 82 degrees

This information is mathematically represented as:

$t+8>82^{\circ}$

7 0

1 year ago

Can someone pls help me

Paladinen [302]

Answer:

y=3x

Step-by-step explanation:

In every collum, the x side increases by one but the y side increases by 3. which means it is multiplying.

0 x 3 = 0

1 x 3 = 3

2 x 3 = 6

3 x 3 = 9

3 0

2 years ago

Read 2 more answers

Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤

MrMuchimi

The average rate of change of a function f(x) in an interval, a < x < b is given by
$\frac{f(b) - f(a)}{b - a}$

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by $\frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4$

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by $\frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3$

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3$

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by $\frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1$

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1$

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by $\frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0$

3 0

3 years ago

Read 2 more answers

Find an equation of the perpendicular bisector of the line segment whose endpoints are given.

REY [17]

Answer:

The equation of the perpendicular bisector line

5 x - 2 y + 31 =0

Step-by-step explanation:

<u><em> Explanation:-</em></u>

Given points are ( 0,1) and (-10,5)

The Midpoint of given two points