<img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7Dx%5E%7B2%7D%20%2B5x%2B%5Cfrac%7B4%7D%7B3%7D%3D2x-%5Cfrac%7B1%7D%7B3%7D%20

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2 years ago

x%5E%7B2%7D" id="TexFormula1" title="\frac{1}{3}x^{2} +5x+\frac{4}{3}=2x-\frac{1}{3} x^{2}" alt="\frac{1}{3}x^{2} +5x+\frac{4}{3}=2x-\frac{1}{3} x^{2}" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Ber [7]2 years ago

5 0

X1= -4, x2= -1/2 is the correct answer

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Z = (4y – 25° n zº 73° y = U

hammer [34]

Answer:

-(2×-4y)

Step-by-step explanation:

calculate it

3 0

2 years ago

The sanchez family goes out for dinner, and the price of the meals is $60. The sales tax on the Meals is 7 percent, and they als

dem82 [27]

Answer:

73.3

Step-by-step explanation:

7%=7÷100=0.07 ×60=4.2

15%=15÷100=0.15×60=9

add you answers to 60 and to should get 73.2

5 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

What is the answer to -1/2+5/4?

babunello [35]

Answer:

3/4

Step-by-step explanation:

6 0

2 years ago

What is the fifth term in the geometric sequence described by this explicit formula y=40×(-2)^(n-1)

cupoosta [38]

Answer:

The fifth term is 620

Step-by-step explanation:

y=40×(-2)^(n-1)

The 5th terms means n=5

y = 40 * (-2) ^ (5-1)

= 40 * (-2) ^4

= 40 * (16)

= 640

5 0

3 years ago