Simplify this equation 3x^2×4x^5y^4

Jan is saving for a skateboard she has saved 30 dollars already which is 20percent of the total price. How much does it cost?

Vesnalui [34]

Answer:

Step-by-step explanation:

so 20% of the total price equals $ 30

let x represent the total price

turn ur percent to a decimal

0.20x = 30

x = 30 / 0.20

x = 150 <=== total price

7 0

3 years ago

Help me on this worksheet and pls show work :)

baherus [9]

What you do to solve this problem you put the faction side-by-side then see what number 16 and six could both go into just multiply 6 times 116 times one and keep on going down to you have the same answer doesn't answer you get is going to be a denominator and then after you get your denominator you do 5 times five and the number you get for you denominator

8 0

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

(No links!) Ill give brainliest and 20+ points to decent answer and this is due today!

Kaylis [27]

Answer:

The answer for number 1 is Bill, his questions is statistical because he is collecting and analyzing data. But Lisa is just asking what their favorite book was.

Step-by-step explanation:

moreover ) Lisa, because there can be many answers to the question.

2) 4.5 points.

4 0

3 years ago

Please Help, I need it for my homework

Naily [24]

Answer:

A)

p1 (0,0) p2 (1,2) given from the graph

slope = (y2-y1)/ (x2-x1 )

slope (m) = 2

y = mx+ b

b =0

y = 2x

b)

x = 0

y = 5

(0,5)

C)

gradient = slope

y = 3x + b

point given (0,12)

y = 3x + 12

5 0

2 years ago

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