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3 years ago

Please help with number 17

Mathematics

1 answer:

melamori03 [73]3 years ago

4 0

Answer:The Answer for the perimeter is 49.

Step-by-step explanation:You just need to add all the sides up to get to 49.

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A large adult elephant weights 24,000 pounds. A baby elephant, on the other hand, weights only 260 pounds. How much more does th

galina1969 [7]

24,000- 260 would equal 23,740 pounds

8 0

3 years ago

Solve for x <br>possibly answer<br><br>12 <br>-7<br>5<br>-6

Natali5045456 [20]

Answer:

12 is the answer. cause 5×12 is 60 and the missing angle is 60 degrees

3 0

3 years ago

Please help I dont understand

Arisa [49]

Answer:

38 degrees

Step-by-step explanation:

By the Law of Cosines,

5² = 3² + 7² - 2(3)(7) cos A

5² - 3² - 7² = -2(3)(7) cos A

-33 = -42 cos A

cos A = 33/42

A = arccos(33/42), which is about 38 degrees.

5 0

2 years ago

If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

4 years ago

What is the opposite reciprocal of 1/-3

Nataly [62]

Answer:

-3

Step-by-step explanation:

hope this helps

5 0

3 years ago

Read 2 more answers