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blsea [12.9K]
3 years ago
8

The sum of 5x2y and (2xy2 + x2y) is

Mathematics
1 answer:
Ludmilka [50]3 years ago
8 0
The sum of all three is 6x^2y + 2xy^2
You might be interested in
A mountain in boone county is eroding and losing elevation at a rate of 5% every millennium. if the current elevation is 1,710 m
aliina [53]

Answer: 1,393 meters tall

Step-by-step explanation:

Plug in 1710 meters for the initial amount, 0.05 for the rate of decrease, and 4 millennia for the time elapsed.

y

= a(1–r)t

y

= 1,710(1–0.05)4 Plug in values

y

= 1,710(0.95)4 Subtract

y

= 1,710(0.81450…) Simplify

y

= 1,392.80568… Multiply

Round to the nearest whole number.

1,392.80568… → 1,393

To the nearest whole number, the mountain will be 1,393 meters tall.

3 0
2 years ago
Ann made a scale drawing of a theater. The scale she used was 1 inch : 7 feet. The stage is
hjlf

Answer:

  • 4 in

Step-by-step explanation:

<u>Use ratios and solve:</u>

  • 1 in ÷ 7 ft = x in ÷ 28 ft
  • x = 28/7 in
  • x = 4 in
4 0
2 years ago
A truck driver transported 748 boxes of goods at the first time. And the number of boxes of goods at the second time decreased t
dedylja [7]

Answer: 25% decrease in number of boxes of goods

Step-by-step explanation:

If you want to find a percent of a number you turn the percent into a decimal

Example:

Step 1

25% of 748

Step 2

0.25*748=561

8 0
2 years ago
We know that Martin is buying some apples at $0.75 each and two dozen oranges
Daniel [21]

Step-by-step explanation:

0.75a + 2(0.90) = 27.60

is the correct answer

8 0
2 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
3 years ago
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