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3 years ago

Distribute -9 (-4+8) please and thank you

Mathematics

1 answer:

qaws [65]3 years ago

5 0

Answer:

36-72

Step-by-step explanation:

all you do is multiply the -9 by the numbers in the parenthesis. If you are multiplying two negatives or two positives then the outcome if positive. If you multiply a positive and a negative the outcome is negative.

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The question I need help in is the middle one. PLEASE I NEED ANDWER ASAP!

antoniya [11.8K]

Answer:

3.25

Step-by-step explanation:

I plugged in 5 for x and got 3.25 when I solved

4 0

4 years ago

What’s the correct answer for this?

kaheart [24]

Answer:

Step-by-step explanation:

Try it -3 squared is 9 -4 squared is 16 , 9+16 = 25 so $x^{2} +y^{2} =25$

7 0

4 years ago

Read 2 more answers

naomi has earned $54 mowing lawns the past two days. she worked 2 1/2 hours yesterday and 4 1/4 hours today. if naomi is pains t

Vaselesa [24]

Answer:

Naomi charges $8 an hour to mow lawns

Step-by-step explanation:

first day // 2 1/2= 2.5 hrs

2nd day // 4 1/4= 4.25 hrs

2.5hrs + 4.25hrs = 54hrs // 6.75hrs = 54hrs

/6.75 /6.75

= 8hrs

Really hope this helps!

8 0

3 years ago

What is the square root of 4 over 225?

exis [7]

Sqrt(4/225)
= 2/15.

The final answer is 2/15~

6 0

3 years ago

The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th

ElenaW [278]

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

$|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}$

By doing so, we find:

$|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}$

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

$|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1$

4 0

3 years ago