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3 years ago

Hellp me on number 3 pls

Mathematics

2 answers:

Y_Kistochka [10]3 years ago

4 0

Answer:

a. +150, b. -15 yards, c. -75 feet, d. +35 degrees, e. -$24.75

Step-by-step explanation:

You are just evaluating the (-) and (+) symbol based off the term being used. If they are gaining something its a (+) if its a loss in any way shape or form it is a (-)

Pavel [41]3 years ago

3 0

Answers:

a) 150
b) -15
c) -75
d) 35
e) -24.75

The values for A and D are positive. The values for B, C, and E are negative.

When you win money, that's a positive thing. Losing money is negative. Being in debt is also negative.

When you're above sea level, then that's a positive height. Below sea level is a negative height. The same applies to temperature as well.

It might help to draw a vertical number line and plot each of these values to see how they're related and how they are sorted. You'll have 150 at the very top, then 35 next on down, followed by -15, -24.75, and finally -75 at the very bottom.

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PtichkaEL [24]

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3 years ago

The measure of angle JKL is 160.

vagabundo [1.1K]

Answer:

x + 160° 95°=360°

x=105°

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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 $[0,1) \cup (3,5)$

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3 years ago

What is the coefficient in this equation 3x +9= -15

VLD [36.1K]

Answer:

Step-by-step explanation:

The coefficient is the number before the variable so 3 would be the coefficient in this equation.

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4 years ago

Suppose you purchase a raffle ticket for $1.00. First prize is $5,000, second prize is $1,000, and third prize is $500. If 500 t

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You should expect to gain 12$

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4 years ago