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rewona [7]
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?
Mathematics
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.

4 0
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>

5 0
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
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