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2 years ago

Why does a rectangle with different side lengths have only 2 lines of symmetry?

Mathematics

1 answer:

ICE Princess25 [194]2 years ago

5 0

Answer:

C. all four angles are congruent, and opposite sides are congruent for a horizontal and vertical line of reflection

Step-by-step explanation:

A. isn't correct because all four sides aren't congruent, only the pair of opposite sides are congruent. Sense only opposite side are congruent you can only reflect the opposite sides onto each other and not diagonally meaning B. and D. are incorrect leaving C. all four angles are congruent, and opposite sides are congruent for a horizontal and vertical line of reflection as the correct answer

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What is the value of a? A.-16 B.-3 C.3 D.16

Igoryamba

Answer:

Which equation do I refer to?

Step-by-step explanation:

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1 year ago

The cost in dollars for removing p percent of pollutants from a river in Smith County is?

ZanzabumX [31]

Ok lets go ahead and see if the following answers can help:
A) if p = 0.2 then C(p) = 157.52
B) if p = 0.95 then C(p) = 753.86
C) smallest p value that gives C(p) > or = 0. This is p = 0
D) largest INTEGER value for C(p) to exist or be >=0. This is p = 99. p = 100 gives 786000/0; does not exist. p > 101 starts giving negative numbers.
I hope this has been of help

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3 years ago

Read 2 more answers

Each week carrie works 40 hours she spend 15% of her time at register

VLD [36.1K]

Please, share the goal of the problem. If Carrie works 40 hours per week and spends 15% of her time at the register, how much time does she NOT spend at the register?

Subtract 15% from 100%; the result is 85%. Now find 85% of 40 hours:

0.85(40 hours) = 34 hours.

Please note that I invented part of this problem. Next time, share all of the info so that someone can help you solve YOUR particular problem.

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3 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

Point W is on line segment VX. given VW=3 and VX=14, determine the length WX

r-ruslan [8.4K]

Answer:11

Step-by-step explanation:14-3=11

4 0

3 years ago