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3 years ago

The equation for this graph has a leading coefficient that is

Mathematics

2 answers:

Neko [114]3 years ago

8 0

Answer:

negative

Step-by-step explanation:

stepladder [879]3 years ago

5 0

Answer:no graph attached

Step-by-step explanation:

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Sue Has 44 Rib bones on display. there are 22 on the right side and 22 on the left on the right of its body on the right side 9/

Murljashka [212]

Answer:

18/44

Step-by-step explanation:

To turn 9/22 into something over 44, we need to figure out what to multiply it by.

44 / 22 = 2.

Multiply 9/22 by 2:

9/22 * 2 = 18/44

4 0

3 years ago

Please help 10 points

m_a_m_a [10]

Answer:

Step-by-step explanation:

A=πr2

= (3.14)(27)2

= 2289.06

5 0

3 years ago

Read 2 more answers

75,575 how do the values of the digit differ in the number?

Vsevolod [243]

Answer:

<u>7</u>5,575 - ten-thousands place; this 7 is worth 70,000

7<u>5</u>,575 - thousands place; this 5 is worth 5,000

75,<u>5</u>75 - hundreds place; this 5 is worth 500

75,5<u>7</u>5 - tens place; this 7 is worth 70

75,57<u>5</u> - ones place; this 5 is worth 5

8 0

3 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

The walls and ceiling of a hotel room are to be painted. The floor is a 14-foot square and the ceiling is 8 feet high. Ignoring

Dennis_Churaev [7]

Answer

112 square feet no

Explanation
I just multiplied I remember solving it that way when I was in middle school

4 0

3 years ago