Ask question

Ask question

All categories

2 years ago

11

A rectangular swimming pool has a length of 50 meters, a width of 25 meters, and a depth of 2 meters. How much paint would be ne

eded to cover the sides and bottom of the pool?

1 answer:

sweet-ann [11.9K]2 years ago

5 0

Answer:

2500m³

Step-by-step explanation:

A = length x width x height

50m x 25m x 2 m

=2500m³

You might be interested in

Please help I have no clue what this is I give brainly :)

nalin [4]

196 + x = 500

Subtract 196 from both sides:
196 on the left is cancelled out, and you’re left with x = 304

:)

6 0

3 years ago

I WILL GIVE BRAINLIEST PLEASE HELP!!!

Rudik [331]

Answer:

1/x+2

Step-by-step explanation:

To find the inverse of the function, try switching the x value with the y value or function value

something like: x=1/g(x)-2

then, just manipulate the equation so that g(x) is by itself again, but this time, since it is an inverse, it's denoted as g-1(x)

x=1/g-1(x)-2

x+2=1/g-1(x)

(g-1(x))(x+2)=1

g-1(x)=1/(x+2)

7 0

3 years ago

Is the following relation a function?<br><br> Yes <br> No

shusha [124]

Yes because none of the x values double

8 0

3 years ago

Please help me , idk the answer

solmaris [256]

Hello!

The answer is 20.

On the left side of the table, you are counting by 1's so of course its going to be 0,1,2,3. Now, on the right side of the table, you are counting by 5's so its going to be 5,10,15,20.

Hope I helped!!!!

-boribaby

3 0

3 years ago

Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!

IgorC [24]

Hey there, hope I can help!

$-4x^2+9y^2+32x+36y-64=0$

$\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \ 9y^2+32x+36y-4x^2=64$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4$
$-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \ -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1$

$Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1$

For me I used
$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1$
$As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}$

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
$\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1$

Hope this helps!

4 0

3 years ago