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3 years ago

14

This building consists of two rectangular prisms-a small space for offices and an attached larger space for warehouse storage. H

ow much space does the building take up?

1 answer:

Naily [24]3 years ago

6 0

I would have to know the space of the two rectangular prisms to answer this correctly

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Which ordered pairs in the form (x, y) are solutions to the equation

Ann [662]

Answer:

A, D

Step-by-step explanation:

Sub the x and y values into the equation. The x value is the first number in the coordinate, and the y value is the second number.

If the left side is equal to the right side, it is a solution.

Example:

4x-5y=24

4(-9)-5(-12)=24

-36+60=24

24=24

Remember: if the left side does not equal the right side, it is not a solution.

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2 years ago

Evaluate each expression.<br> 8/7 + 4/7

Galina-37 [17]

Answer:

<u>1 4/7</u>

Step-by-step explanation:

8/7 + 4/7 = 12/7. Simplified = <u>1 4/7</u>

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2 years ago

Read 2 more answers

The number 6 is a solution to which of the following inequalities?

Paladinen [302]

Answer:

21 + x < 28

Step-by-step explanation:

27 is in fact greater than 28 if you where to input 6 into the x spot, this is the only one that's a true expression.

8 0

3 years ago

Four friends went scuba diving today. Ali dove 70 feet, Tim went down 50 feet, Carl dove 65 feet, and Brenda reached 48 feet bel

ss7ja [257]

Ali dove the furthest

7 0

3 years ago

A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,

guajiro [1.7K]

We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation $A=45e^{-0.05t}$ .

To find the time it will take for the area of the glacier to decrease to 15 square kilo-meters, we will equate $A=15$ and solve for t as:

$15=45e^{-0.05t}$

$\frac{15}{45}=\frac{45e^{-0.05t}}{45}$

$\frac{1}{3}=e^{-0.05t}$

Now we will switch sides:

$e^{-0.05t}=\frac{1}{3}$

Let us take natural log on both sides of equation.

$\text{ln}(e^{-0.05t})=\text{ln}(\frac{1}{3})$

Using natural log property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$-0.05t\cdot \text{ln}(e)=\text{ln}(\frac{1}{3})$

$-0.05t\cdot (1)=\text{ln}(\frac{1}{3})$

$-0.05t=\text{ln}(\frac{1}{3})$

$t=\frac{\text{ln}(\frac{1}{3})}{-0.05}$

$t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-0.05\cdot 100}$

$t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-5}$

$t=-\text{ln}(\frac{1}{3})\cdot 20$

$t=-(\text{ln}(1)-\text{ln}(3))\cdot 20$

$t=-(0-\text{ln}(3))\cdot 20$

$t=20\text{ln}(3)$

Therefore, it will take $20\text{ln}(3)$ years for area of the glacier to decrease to 15 square kilo-meters.

6 0

2 years ago