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DENIUS [597]
3 years ago
8

Are the following figures similar? Rectangles ABCD and EFGH are shown. AB equals 5. BC equals 25. EF equals 3. FG equals 15. Yes

; the corresponding angles are congruent No; the corresponding angles are not congruent Yes; the corresponding sides are proportional No; the corresponding sides are not proportional
Mathematics
2 answers:
Tatiana [17]3 years ago
8 0

Answer:

Yes; the corresponding sides are proportional

Step-by-step explanation:

If two figures are similar, the lengths of their sides are proportional.  This means if we set up a proportion with the sides we are given and cross-multiply, we get a true statement at the end of it.

Using the similarity statement ABCD~EFGH, we will compare AB to EF and BC to FG:

5/3 = 25/15

Cross multiply:

5(15) = 3(25)

75 = 75

We got a true statement, so the sides are proportional.

svlad2 [7]3 years ago
7 0

The correct answer among all the choices is C) Yes; the corresponding sides are proportional. This is the correct answer because the corresponding sides have a similar relationship of 1:5. We can see this relationship when dividing 25 by 5 & dividing 15 by 3, since their both equal to 5, we know the figures are similar. I hope I answered this question to your satisfaction. Have a good day!

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If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by
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Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by y = 70t-16t^2.

To find : The average velocity for the time period beginning when t = 2 and lasting.  a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

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a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}

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b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}

(\text{Average velocity})_{0.01\ s}=7536\ ft/s

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}  

(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}

(\text{Average velocity})_{0.001\ s}=75936\ ft/s

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