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3 years ago

ANSWER PLEASE! Pic attached.

Mathematics

1 answer:

olasank [31]3 years ago

8 0

Ah, Micah, that rat.

The answer is B, I took the test and that was correct.

If you break the equation up into systems, you can see that they intersect at 0.297, which would be the solution to the systems.

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The endpoints of the diameter of a circle are (0,0 and (0,-6). what is the area of the circle?

makvit [3.9K]

Answer: $9\pi$

Step-by-step explanation:

DIAMETER length = 6

radius length = 3

Area of the circle = $\pi r^{2} = 9\pi$

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2 years ago

What is the perimeter of this quadrilateral?<br> (5,5)<br> (2, 4)<br> (4, 1)<br> (6, 1)

lbvjy [14]

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{5}~,~\stackrel{y_1}{5})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{4}) ~\hfill AB=\sqrt{[ 2- 5]^2 + [ 4- 5]^2} \\\\\\ AB=\sqrt{(-3)^2+(-1)^2}\implies \boxed{AB=\sqrt{10}} \\\\[-0.35em] ~\dotfill\\\\ B(\stackrel{x_1}{2}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{1}) ~\hfill BC=\sqrt{[ 4- 2]^2 + [ 1- 4]^2}$

$BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\qquad D(\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) ~\hfill CD=\sqrt{[ 6- 4]^2 + [ 1- 1]^2} \\\\\\ CD=\sqrt{2^2+0^2}\implies \boxed{CD=2} \\\\[-0.35em] ~\dotfill\\\\ D(\stackrel{x_1}{6}~,~\stackrel{y_1}{1})\qquad A(\stackrel{x_2}{5}~,~\stackrel{y_2}{5}) ~\hfill DA=\sqrt{[ 5- 6]^2 + [ 5- 1]^2}$

$DA=\sqrt{(-1)^2+4^2}\implies \boxed{DA=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{\sqrt{10}~~ + ~~\sqrt{13}~~ + ~~2~~ + ~~\sqrt{17}}~~ \approx ~~ 12.89$

8 0

3 years ago

What is the area of the polygon??

muminat

B 95.5 hope it helps

6 0

3 years ago

Which expression is not equal to one fourth of 52?

Zina [86]

Answer:

B) 4% of 52

Step-by-step explanation:

A) 0.25 x 52 = 13

B) 4% of 52 = 0.04 x 42 = 2.08

C) 52 ÷ 4 = 13

D) 52/4 = 13

4 0

3 years ago

Read 2 more answers

The product of the first and third of three consecutive integers is 3 more than 3 times the second integer. Find the integers.

puteri [66]

Let:

x = the middle integer

Then the 3 consecutive integers are:

x - 1, x, x + 1

The product of the first and the third integer:

(x - 1)*(x + 1) = x^2 - 1

is equal to:

3 more than 3 times the second integer:

3x + 3

So:

x^2 - 1 = 3x + 3

Solve:

x^2 - 3x - 4 = 0

(x-4)(x+1) = 0

x = -1 and 4

Since x is the middle integer, we have two sets of solutions:

3,4,5 and -2,-1,0

4 0

3 years ago