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Kobotan [32]
3 years ago
9

What is the length of the midsegment of a trapezoid with bases of length 16 and 22

Mathematics
1 answer:
Debora [2.8K]3 years ago
6 0
1/2(16+22)
= 1/2 (38)
= 19
the length if the midsegment is 19







I'm pretty sure...
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3 years ago
Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds a
kirill115 [55]

Answer:

12 feet per second.

Step-by-step explanation:

Please consider the complete question.

Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds after she jumps for the bone can be representedh(t)=-16t^2+20t.

What is Spot's average rate of ascent, in feet per second, from the time she jumps into the air to the time she catches the bone at t=1/2?  

We will use average rate of change formula to solve our given problem.

\text{Average rate of change}=\frac{f(b)-f(a)}{b-a}

\text{Average rate of change}=\frac{h(\frac{1}{2})-h(0)}{\frac{1}{2}-0}

\text{Average rate of change}=\frac{-16\cdot(\frac{1}{2})^2+20\cdot \frac{1}{2}-(-16\cdot(0)^2+20\cdot (0))}{\frac{1}{2}-0}

\text{Average rate of change}=\frac{-16\cdot\frac{1}{4}+10-(0)}{\frac{1}{2}}

\text{Average rate of change}=\frac{-4+10}{\frac{1}{2}}

\text{Average rate of change}=\frac{6}{\frac{1}{2}}

\text{Average rate of change}=\frac{2\cdot 6}{1}

\text{Average rate of change}=12

Therefore, Spot's average rate of ascent is 12 feet per second.

3 0
3 years ago
BRAINLIESTTT ASAP!! PLEASE HELP ME :)
Lostsunrise [7]

Answer:

See below  

Step-by-step explanation:

(a) Field lines

A negatively charged particle has an electric field associated with it.

The field lines spread out radially from the centre of the point. They are represented by arrows pointing in the direction that a positive charge would move if it were in the field.

Opposite charges attract, so the field lines point toward the centre of the particle.

For an isolated negative particle, the field lines would look like those in Figure 1 below.

If two negative charges are near each other, as in Figure 2, the field lines still point to the centre of charge.

A positive charge approaching from the left is attracted to both charges, but it moves to the closer particle on the left.

We can make a similar statement about appositive charge approaching from the left.

Thus, there are few field lines in the region between the two particles.

(b) Coulomb's Law

The formula for Coulomb's law is

F = (kq₁q₂)/r²

It shows that the force varies inversely as the square of the distance between the charges.

Thus, the force between the charges decreases rapidly as they move further apart.

5 0
3 years ago
15Points!!<br> Factor completely.<br><br><br> 12x² +28x
Zigmanuir [339]

Answer:

4x(3x+7)

Step-by-step explanation:

12x^2+28x\\12xx+28x\\\\4\cdot \:3xx+4\cdot \:7x\\4x\left(3x+7\right)

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What is it<br><br> −4(3/2x−12)=−15
Greeley [361]

Answer:

x = 21/2

Step-by-step explanation:

Step 1: Write equation

-4(3/2x - 12) = -15

Step 2: Solve for <em>x</em>

  1. Distribute -4: -6x + 48 = -15
  2. Subtract 48 on both sides: -6x = -63
  3. Divide both sides by -6: x = 21/2
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3 years ago
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