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2 years ago

- The position of an object moving in a straight line

Mathematics

1 answer:

Misha Larkins [42]2 years ago

4 0

Answer:

s(t) = 2t^2-3t

19 = 2t^2-3t

2t^2-3t-19=0

t = [3 +- sqrt(9-4*2*-19)]/4

t = [3 +- sqrt(161)]/4

t = [3 +- 12.6886]/4

Positive answer:

t = [3+12.6886]/4

t = 3.9221 seconds

Step-by-step explanation:

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What is pie? explain

Kaylis [27]

Answer: Do you mean a pie chart or a dessert pie? anyway i will answer for the pie chart

Step-by-step explanation:

pie charts are used in data handling and they are circular divided up into different segments which each represent a value or a different size

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2 years ago

Can someone say what 45-49 is with regrouping? thanks!

muminat

Answer:

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Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

8 = 4 w divided by 3

olya-2409 [2.1K]

8 = 4w/3

8 = 4w/3
×3 = 4w×3
24= 4w ( the two 3s cross each other out)
÷4 = ÷4
8 = w ( the two 4s cross each other out)

6 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

In the diagram below, C is the midpoint of AB. If AC is 4 centimeters, what is

Strike441 [17]

Answer:

The answer is B. 4cm

Step-by-step explanation:

Since it is the midpoint AC is always going to be the same as CB if C is the midpoint.

6 0

2 years ago