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3 years ago

Please answer for brainlesit

Mathematics

2 answers:

lorasvet [3.4K]3 years ago

5 0

(-3, -2) and here's why. The first quadrant is (+, +) in the top right.

The second quadrant is (-, +)

The third quadrant is (-, -) where the auditorium is located

And lastly, the fourth quadrant (+, -)

The English room is in the same quadrant as the auditorium, being quadrant 3

Have a brainly day :)

Brums [2.3K]3 years ago

4 0

Answer:

(-3, -2) that is the answer please give me brainiest

Step-by-step explanation:

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Write the equation of the line that is the perpendicular bisector of the segment with endpoints (4, 1) and (2, -5)

Sedbober [7]

Answer: $\bold{y=-\dfrac{1}{3}x-1}$

<u>Step-by-step explanation:</u>

(4, 1) & (2, -5)

First, find the slope (m) and then the perpendicular (opposite reciprocal) slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\\m=\dfrac{-5-1}{2-4} = \dfrac{-6}{-2}=3\quad \rightarrow \quad m_{\perp}=-\dfrac{1}{3}\\$

Next, find the midpoint of (4, 1) and (2, -5):

$Midpoint=\bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)\\\\\\.\qquad \qquad=\bigg(\dfrac{4+2}{2},\dfrac{1-5}{2}\bigg)\\\\\\.\qquad \qquad=\bigg(\dfrac{6}{2},\dfrac{-4}{2}\bigg)\\\\\\.\qquad \qquad=(3, -2)$

Lastly, input the perpendicular slope and the midpoint into the Point-Slope formula to find the equation of the line:

$y - y_1 = m_{\perp}(x - x_1)\\\\y - (-2) = -\dfrac{1}{3}(x - 3)\\\\y + 2=-\dfrac{1}{3}x +1\\\\y =-\dfrac{1}{3}x - 1\\$

6 0

2 years ago

HELP ASAP WILL MARK BRAINLIEST!<br><br> What is the length of TW?

love history [14]

Answer:

Step-by-step explanation:

You are given the hypotenuse and one leg of a right triangle, and are being asked to find the other leg. The Pythagorean theorem applies.

TW² + TY² = WY²

TW² = 17² - 15² = 64

TW = √64

TW = 8

3 0

3 years ago

Read 2 more answers

Select the function that's represented in the graph. Question 5 options: A) ƒ(x) = x2 – 3 B) ƒ(x) = –1∕2x2 – 3 C) ƒ(x) = –x2 – 3

Neporo4naja [7]

Answer:

Step-by-step explanation:

hello,

because of the end behaviour the constant in $x^2$ should be positive so we have a) or d)

f(0)=-3 in both cases

for A) f(x)=

$x^2-3=(x-\sqrt{3})(x+\sqrt{3})$

so f(x)=0 for $x=\sqrt{3} \ or \ x = -\sqrt{3}$

so the correct answer is A)

hope this helps

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2 years ago

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Ivahew [28]

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