11. Use the information to answer the questions.

Help please anyone! Brainliest reward it counting on it!

Viefleur [7K]

X1 + x2 / 2 , y1 + y2 / 2

8 + x2 /2 = 6

So x is 4

9 + y2 /2 = 6

So y is 3.

Answer is B ( 4, 3)

Please Mark Brainliest if it helped

4 0

3 years ago

Select all the statements you agree with.

wolverine [178]

The statements that I agree with, with respect to a parallelogram are:

A parallelogram has six sides.
Opposite sides of a parallelogram are parallel.
All angles of a parallelogram have the same measure.
All sides of a parallelogram have the same length.

<h3>What is a parallelogram?</h3>

A parallelogram can be described as the quadrilateral that do contains two pairs with respect to their parallel sides.

It should be noted that the sides that can be found in the parallelogram are usually of the same length and there the sides that can be attributed to a parallelogram is usually 6.

Therefore, the option 1,2,4,5 are correct.

Learn more about parallelogram at:

brainly.com/question/970600

#SPJ1

6 0

1 year ago

What is the value of the expression when n=3 -2n(5+n-8-3n)

igor_vitrenko [27]

-2(3)(5+3-8-3(3))
=-6(5+3-8-9)
=-30-18+48+54
=54

4 0

3 years ago

Read 2 more answers

A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?

kirza4 [7]

Answer:

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

$f(x)=\frac{1}{2^{2x}}$ (1)

Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

You first derivative the function f(x):

$\frac{df}{dx}=\frac{d}{dx}[\frac{1}{2^{2x}}]$ (2)

To solve this derivative you use the following derivative formula:

$\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}$

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

$\frac{d}{dx}[2^{-2x}]=2^{-2x}(ln2)(-2)$

You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

$-2(ln2)2^{-2x}=-1$

Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

$-2(ln2)2^{-2a}=-1 \\\\2^{2a}=2(ln2)=1.386\\\\log_22^{2a}=log_2(1.386)\\\\2a=\frac{log(1.386)}{log(2)}\\\\a=0.235$

With this value you calculate f(a):

$f(a)=\frac{1}{2^{2(0.235)}}=0.721$

Next, you use the general equation of line:

$y-y_o=m(x-x_o)$

for xo = a = 0.235 and yo = f(a) = 0.721:

$y-0.721=(-1)(x-0.235)\\\\y=-x+0.956$

The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

$0=-x+0.956\\\\x=0.956$

hence, the x-intercept of the tangent line is 0.956

5 0

3 years ago

Consider the following points on cartesian plane. A (1;2) BC3; 1) and CG-3; 4) prove that Points A, B and C are collinear.

Arlecino [84]

Step-by-step explanation:

Collinear points have the same gradient

find the gradient(slope) between AB first

$= \frac{1 - 2}{3 - 1} \\ = \frac{ - 1}{2}$

Find the one for BC

$= \frac{4 - 1}{ - 3 - 3} \\ = \frac{3}{ - 6} \\ = \frac{ - 1}{2}$

Find the gradient between AC now

$= \frac{4 - 2}{ - 3 - 1} \\ = \frac{2}{ - 4} \\ = \frac{ - 1}{2}$

YOUR LAST STATEMENT NOW WILL BE

POINTS A,B AND C ARE COLLINEAR SINCE THE GRADIENTS BETWEEN THEM IS THE SAME!!!

5 0

1 year ago