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3 years ago

10

What is the surface area of the box if it is scaled up by a factor of 10? The surface area is: 736 cm. Lenght: 12 cm. Width: 4 c

m. Height: 20 cm.

1 answer:

grandymaker [24]3 years ago

6 0

739*12*4*20=71,000 divded by 2=36,000

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Real life situation where an estimate makes sense

harina [27]

I'm in a hurry to get home.
My car is almost out of gas.
I pull into a gas station.
I have $10 in my pocket.
Gas is selling for $2.36⁹ per gallon.
How many gallons can I buy.

=================================

Even better:

My car gets 25 miles to the gallon.
How many miles can I buy ?

6 0

3 years ago

Help me out 3 i need some help :(

Alenkinab [10]

Answer:

The first one

Step-by-step explanation:

Since only the first session is $10, it wouldn't be 10x.

The second session is $5, and it will never be $10 again, so $5 sessions are unlimited which would be 5x.

So the answer is y = 10 + 5x

(sorry if i didnt explain well)

8 0

3 years ago

hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and

Mariulka [41]

Given:

The equation is,

$2\log _3x-\log _3(x-2)=2$

Explanation:

Simplify the equation by using logarthimic property.

$\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{ \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}$

Simplify further.

$\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}$

Solve the quadratic equation for x.

$\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}$

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

$\begin{gathered} x-6=0 \\ x=6 \end{gathered}$

For (x - 3) = 0,

$\begin{gathered} x-3=0 \\ x=3 \end{gathered}$

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

$\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}$

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

$\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}$

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0

1 year ago

Aika has 294 one-foot unit cubes. He use them to build a rectangular prism. He has 6 cubes left over. Which could be the dimensi

SSSSS [86.1K]

If he has 6 left over it has to be 288 cubic feet so the dimensions could be 12x12x2

7 0

3 years ago

Suppose that profit for a particular product is calculated using the linear equation: Profit = 20S + 3D. Which of the following

uranmaximum [27]

Answer:

b. S = 405, D = 0

Step-by-step explanation:

We have been given that profit for a particular product is calculated using the linear equation: $\text{Profit}=20S+3D$ . We are asked to choose the combinations of S and D that would yield a maximum profit.

To solve our given problem, we will substitute given values of S and D in the profit function one by one.

a. S = 0, D = 0

$\text{Profit}=20S+3D$

$\text{Profit}=20(0)+3(0)$

$\text{Profit}=0$

b. S = 405, D = 0

$\text{Profit}=20S+3D$

$\text{Profit}=20(405)+3(0)$

$\text{Profit}=8100+0$

$\text{Profit}=8100$

c. S = 0, D = 299

$\text{Profit}=20S+3D$

$\text{Profit}=20(0)+3(299)$

$\text{Profit}=0+897$

$\text{Profit}=897$

d. S = 182, D = 145

$\text{Profit}=20S+3D$

$\text{Profit}=20(182)+3(145)$

$\text{Profit}=3640+435$

$\text{Profit}=4075$

Since the combination S = 405, D = 0 gives the maximum profit ($8100), therefore, option 'b' is the correct choice.

7 0

3 years ago