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Miguel glues a ribbon border around the edges of a 5 - inch by 8 - inch picture to create a frame. What is the total length of r

Marat540 [252]

26 inches. Technically, since this is a perimeter problem, you need to add each length to itself and then the others. If this is confusing, you essentially need to do 5+5+8+8=26 because you have two of the fives and two eights. Perimeter is much easier than area, I promise you that

3 0

3 years ago

What is the equation for the values in the table

devlian [24]

Step-by-step explanation:

Taking any two points through which the line passes;(-10,-3) and (-5,-1)

The equation of the line is given by

$\frac{y2 - y1}{x2 - x1} = \frac{y - y1}{x - x1}$

$=> \frac{ - 1 - ( - 3)}{ - 5 - ( - 10)} = \frac{y - ( - 3)}{x - ( - 10)} = \frac{y + 3}{x + 10}$

$=> \frac{ - 1 + 3}{ - 5 + 10} = \frac{y + 3}{x + 10}$

$=> \frac{2}{5} = \frac{y + 3}{x + 10}$

$=> 5(y + 3) = 2(x + 10)$

=> 5y+15 = 2x+20

=> 5y =2x+5

$=> y= \frac{2}{5} x + 1$

option C

8 0

3 years ago

If michelle rollerblades around a circular track with a radius of 80 meters, how far does she skate? use 3.14 for pi. round to t

Andrew [12]

I believe it’s A but not quite sure

3 0

3 years ago

Read 2 more answers

What is 13200 as a ratio?

tamaranim1 [39]

Answer:

13200n: 4 is 600

Step-by-step explanation:

7 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

3 years ago