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2 years ago

Given B(-4,-6), under which reflection is B(4,-6)?

1 answer:

8 0

Well the one given is in the third quadrant and the other one, B(4,-6) is in the second quadrant. And it's reflected in the x-axis... Is that what you're looking for?

REALLY hope this helps :)

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Plot the points A(9, 11) and B(–3, –5). Find midpoint M of AB. Then show that AM = MB and AM + MB =AB

Alex_Xolod [135]

Answer:

The midpoint is (3, 3).

Step-by-step explanation:

We are given the two points A(9, 11) and B(-3, -5).

The midpoint is given by:

$\displaystyle M=\Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big)$

So:

$\displaystyle M = \Big( \frac{9+(-3) }{2}, \frac{ 11+(-5) }{2} \Big) = (3,3)$

The midpoint is (3, 3).

We want to show that AM = MB.

We can use the distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

The distance between A(9, 11) and M(3, 3) will then be:

$AM=\sqrt{(9-3)^2+(11-3)^2}=\sqrt{6^2+8^2}=\sqrt{100}=10$

And the distance between B(-3, -5) and M(3, 3) will be:

$MB = \sqrt{ (3-(-3))^2 + (3-(-5))^2 } = \sqrt{(6)^2+(8)^2} = \sqrt{ 100 } = 10$

So, AM = MB = 10.

Since AM = MB = 10, AM + MB = 10 + 10 = 20.

So, we want to prove that AB = 20.

By the distance formula:

$AB=\sqrt{(9-(-3))^2+(11-(-5))^2}=\sqrt{12^2+16^2}}=\sqrt{400}=20\stackrel{\checkmark}{=}20$

4 0

2 years ago

Write an exponential function in the form y=ab^xy=ab x that goes through points (0, 20)(0,20) and (2, 720)(2,720).

Nataliya [291]

Answer:

The exponential function that goes through $(x_{1},y_{1}) = (0,20)$ and $(x_{2}, y_{2}) = (2,720)$ is $y = 20\cdot 6^{x}$ .

Step-by-step explanation:

Let be an exponential function of the form $y = a\cdot b^{x}$ that goes through $(x_{1},y_{1}) = (0,20)$ and $(x_{2}, y_{2}) = (2,720)$ . Then, we have the following system of equations: