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2 years ago

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Caroline scored 8 more points than three times the number of points as Kamila. If Kamila scores 8 points in her next try, she wi

ll have scored 40 points for the season.
Question 1
Part A

1 answer:

nlexa [21]2 years ago

8 0

Answer: 32

step by step explanation: 40 - 8 = 32

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At dinner, the meal cost $22 and a $1.87 sales tax was added to the bill. What would be the total amount if the meal cost $80.

Helen [10]

1.87 is 8.5% of 22 so 8.5% of 80 is 0.085 x 80 so $6.80

3 0

2 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

2 years ago

dlinn [17]

Step-by-step explanation:

I hope you have a good day

5 0

3 years ago

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A new truck cost $32,000. The truck's value

Kisachek [45]

Answer: $17,600

Step-by-step explanation:

find the value of 9% of 32,000 (Divide). Multiply that by 100. multiply by 5. Subtract the product by 32,000

7 0

1 year ago

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Explain why sin^-1(sin(3pi/4))does not equal 3pi/4 when y=sin x and y=sin^-1 x are inverses.

Fofino [41]

Because they're not inverses, not exactly. $\sin x$ is not invertible on its entire domain because it's not one-to-one. There are infinitely many values of $x$ such that $\sin x=0$ , for example.

The standard function $\sin^{-1}x$ has a domain of $-1\le x\le1$ and outputs values between $-\dfrac\pi2$ and $\dfrac\pi2$ . This means that its inverse, $\sin x$ , is indeed its inverse as long as $-\dfrac\pi2\le x\le\dfrac\pi2$ .

$\dfrac{3\pi}4$ is larger than $\dfrac\pi2$ and thus does not fall in the "invertible part" of the domain of $\sin x$ . We have

$\sin\dfrac{3\pi}4=\dfrac1{\sqrt2}$

which is a value between -1 and 1, so that

$\sin^{-1}\left(\sin\dfrac{3\pi}4\right)=\sin^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4$

If we wanted to recover $\dfrac{3\pi}4$ , we'd have to redefine $\sin^{-1}x$ or define a new inverse function that works on a different branch of the domain of $\sin x$ .

7 0

3 years ago