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3 years ago

12

A cube has the width of 6 cm.What is the volume ofcthe cube?

2 answers:

Maurinko [17]3 years ago

8 0

Answer:

216

Step-by-step explanation:

l x h x w\

or

length times height times width

Svet_ta [14]3 years ago

4 0

Step-by-step explanation:

Yᴏᴜʀ ᴀɴsᴡᴇʀ ɪs ɪɴ ᴛʜᴇ ᴀᴛᴛᴀᴄʜᴍᴇɴᴛ...

Hᴏᴘᴇ ɪᴛ ʜᴇʟᴘs ʏᴏᴜ

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Can someone find the surface area please

lozanna [386]

To find the surface area of any object, the first step that you must do is find all shapes that are present in the object.

Here we have 2 triangles and 3 rectangles.

Knowing this, simply find the area of each shape.

Since the triangles are the same, one is facing up and the other down, they both would be 15.6 cm^2.

The rectangles would be 54 cm^2. as there are 3 ( 6 x 3 cm) ones.

Add the values up to have the surface area.

It’s 85.2 cm^2.

3 0

3 years ago

Read 2 more answers

4.00 is what percent of 50.00

DiKsa [7]

$Make\ a\ fraction\ \frac{4}{50}\ and\ multiply\ by\ 100\%\\\\
\frac{4}{50}*100\%=\frac{400}{50}\%=8\%\\\\
4\ is\ 8\%\ of\ 50.$

7 0

3 years ago

A number rounded to the nearest ten is 550. When it is rounded to the nestest hundred,the number become 600 which could it ve?

amid [387]

Answer:

550-554

Step-by-step explanation:

550 is rounded to 550, and also rounded to 600. Same with all the other numbers. :)

8 0

3 years ago

In what quadrant does the angle 9pi/ 5 terminate?

aev [14]

the angle 4pi/5 in quadrent 2

3 0

3 years ago

Read 2 more answers

What is the range of the equation

a_sh-v [17]

The range of the equation is $y>2$

Explanation:

The given equation is $y=2(4)^{x+3}+2$

We need to determine the range of the equation.

<u>Range:</u>

The range of the function is the set of all dependent y - values for which the function is well defined.

Let us simplify the equation.

Thus, we have;

$y=2 \cdot 4^{x+3}+2$

This can be written as $y=2^{1+2(x+3)}+2$

Now, we shall determine the range.

Let us interchange the variables x and y.

Thus, we have;

$x=2^{1+2(y+3)}+2$

Solving for y, we get;

$x-2=2^{1+2(y+3)}$

Applying the log rule, if f(x) = g(x) then $\ln (f(x))=\ln (g(x))$ , then, we get;

$\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)$

Simplifying, we get;

$(1+2(y+3)) \ln (2)=\ln (x-2)$

Dividing both sides by $\ln (2)$ , we have;

$2 y+7=\frac{\ln (x-2)}{\ln (2)}$

Subtracting 7 from both sides of the equation, we have;

$2 y=\frac{\ln (x-2)}{\ln (2)}-7$

Dividing both sides by 2, we get;

$y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}$

Let us find the positive values for logs.

Thus, we have,;

$x-2>0$

$x>2$

The function domain is $x>2$

By combining the intervals, the range becomes $y>2$

Hence, the range of the equation is $y>2$

7 0

3 years ago