I need help pleaseee its due in 10 min

Mathematics

2 answers:

Alborosie2 years ago

6 0

What do you looking for?

MrRa [10]2 years ago

3 0

Answer:

What do you need help with?

Step-by-step explanation:

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PLEASE HELP ME!! FIND THE VOLUME OF X

sineoko [7]

Answer:

x = 6

Step-by-step explanation:

So thee cube is just x^3

The cube root of 216 is 6

7 0

3 years ago

Help pls!! Find the area

Dima020 [189]

Answer:

17.5 for perpendicular triangle

Step-by-step explanation:

A= 1/2 ( b*h )

A= 1/2 ( 7*5)

A= 1/2 ( 35)

A= 17.5

8 0

2 years ago

$85 Restaurant Bill; 15% Tip

svp [43]

Answer:

97.75

Step-by-step explanation:

15% of 85 is 12.75

12.75 + 85 = 97.75

Hope this helps have a nice day

8 0

2 years ago

What is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)?

Rasek [7]

The side lengths of triangle are 6 units, 8 units and 10 units.

<u>SOLUTION: </u>

Given that, we have to find what is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)

We know that, distance between two points $P\left(x_{1}, y_{1}\right) \text { and } Q\left(x_{2}, y_{2}\right)$ is given by

$P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now,

$\begin{array}{l}{\text { Distance between }(-5,-1) \text { and }(-5,5)=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {\qquad \begin{array}{l}{=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(0)^{2}+(5+1)^{2}}=\sqrt{(6)^{2}}=6} \\\\ {=\sqrt{(-5)^{2}+(5+1)^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(-8)^{2}+(5+1)^{2}}=\sqrt{64+36}=\sqrt{100}=10} \\\\ {=\sqrt{(3-1)^{2}+(-1-(-1))^{2}}} \\\\ {=\sqrt{(5+3)^{2}+(0)^{2}}=\sqrt{(8)^{2}}=8}\end{array}}\end{array}$