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3 years ago

Help me with my homework please

Mathematics

1 answer:

KatRina [158]3 years ago

3 0

Answer:

Exact Form: $\frac{13}{4}$

Decimal Form: 3.25

Mixed Number Form: $3\frac{1}{4}$

Step-by-step explanation:

hope it helps ;)

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What is the length (magnitude) of the vector (7,-2)?

Ulleksa [173]

Answer:

d. sqrt 53

Step-by-step explanation:

7 0

3 years ago

Please answer with everything needed i appreciate everyones help

yawa3891 [41]

Recheck the answer with your teacher.

4 0

2 years ago

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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

Need this worked out please so I can understand it

Afina-wow [57]

9514 1404 393

Answer:

a) 12

b) y +5 = 12(x +2)

Step-by-step explanation:

The slope of a function is its derivative.

a) The slope of y = x^3 +3 at any point is y' = 3x^2. So, at the point where x=-2, the slope of the curve is ...

m = 3(-2)^2 = 12 . . . . slope at (-2, -5)

b) The equation of the tangent line is most easily written using the point-slope form of the equation of a line.

y -k = m(x -h) . . . . . line with slope m through point (h, k)

y +5 = 12(x +2) . . . . line with slope 12 through point (-2, -5)

3 0

3 years ago

The function C(t) = 400 + 30t models the number of classrooms, C. in the town of Sirap, t years from

maks197457 [2]

Answer:

1) S(t) = C(t) × D(t)

2) S(t) = (400 + 30t)(25 + t)

Step-by-step explanation:

The function C(t) = 400 + 30t ........... (1), models the number of classrooms, C. in the town of Sirap, t years from now.

The function D(t) = 25 + t ......... (2) models the number of students per classroom, D, t years from now.

Then if S(t) represents the number of students in Sirap's school system t years from now, then, we can write the relation

1) S(t) = C(t) × D(t) (Answer)

2) Hence, the formula of S(t) in terms if t is given by

S(t) = (400 + 30t)(25 + t) (Answer)

7 0

3 years ago

Read 2 more answers