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leva [86]
1 year ago
15

A runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her

big push to the finish ?
Mathematics
1 answer:
bezimeni [28]1 year ago
4 0
We know, acceleration = change in speed / time
a = (3.5 - 3.1) / 15
a = 0.4 / 15
a = 0.026 m/s²

In short, Your Answer would be: 0.026 m/s²

Hope this helps!
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Sam has some red and yellow cubes. She has 20 cubes in total. She has 8 more yellow cubes than red ones. How many red cubes does
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Answer:

12

Step-by-step explanation:

because if she has 8 more red cubes that yellow than that would mean it would become 12

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What is x if the volume of the box is 70 in3
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4 times 70 divested byv3
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Find the volume of the composite solid. Round to the nearest tenths. Radius:2 height:4
yKpoI14uk [10]

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6 0
2 years ago
Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1
astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

x^{2} - 24 = 1

Given that resulting expression is a second order polynomial of the form x^{2} - a^{2}, there are two real and distinct solutions. Roots of the expression are:

x_{1} = -5 and x_{2} = 5.

Now, it is also required to determine which part of the interval (x_{1}, x_{2}) is equal to a number greater than zero (positive). That is:

x^{2} - 24 > 0

x^{2} > 24

x < -4.899 and x > 4.899.

Therefore, exists two sub-intervals: [-5, -4.899] and \left[4.899,5\right]. Besides, x^{2} - 24 > y = 1 in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx

A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx

A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}

A = 2.525 -2.474+9.798 + 2.525 - 2.474

A = 9.9\,units^{2}

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

4 0
1 year ago
A biologist is studying the growth of a particular species of algae. She writes the following equation to show the radius of the
vitfil [10]

Let's solve for d.

fd=(7)(1.06)d

Step 1: Add -7.42d to both sides.

df+−7.42d=7.42d+−7.42d

df−7.42d=0

Step 2: Factor out variable d.

d(f−7.42)=0

Step 3: Divide both sides by f-7.42.

d(f−7.42)f−7.42=0f−7.42

d=0f−7.42

Answer:

d=0f−7.42

PLEASE MARK ME AS BRAINLIEST

4 0
1 year ago
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