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2 years ago

Cynthia typed 372 words in 6 minutes what's her typing rate in words per minute

Mathematics

1 answer:

VMariaS [17]2 years ago

5 0

Answer:

x = 62 words

Explanation:

→ We write the data problems.

372 words .............6 minutes

→ We must find: What's Cynthia typing rate in words per minute?

372 words .............6 minutes

x words ..................1 minute

x = 372 words * 1 minute/ 6 minutes

x = 62 words

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

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3 years ago

Wes had 4.75 in quarters and dimes if he had 25 coins how many of each did he have?

Verizon [17]

Answer:it would be 15 quarters and 10 dimes

Step-by-step explanation:

15 quarters is 3.75 and 10 dimes is $1, add them up you get $ 4.75

5 0

3 years ago

Read 2 more answers

Quadrilateral DEFG has vertices D(−2,4) , E(4,7) , F(10,3) , and G(8,0) .

fredd [130]

A rotation 270° counterclockwise about the origin is the same as rotation 90° clockwise about the origin and has a rule:

(x,y)→(y,-x).

Then:

D(−2,4)→D'(4,2)
E(4,7)→E'(7,-4)
F(10,3)→F'(3,-10)
G(8,0)→G'(0,-8)

Answer: the coordinates of vertices of quadrilateral D′E′F′G′ are D'(4,2), E'(7,-4), F'(3,-10), G'(0,-8).

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2 years ago

PLZ HELP ME NEED THIS FAST WILL GIVE BRAINLIEST

OleMash [197]

Answer: l = 60 ft and w = 30 ft

P = 2(l + w)

180 = 2(l + w)

Let x = width of his yard

Let 2x = length of his yard

$180=2(2x+x)$

$\frac{180}{2}=\frac{2(2x+x)}{2}$

$90=3x$

$\frac{90}{3} =\frac{3x}{3}$

$x=30ft$

l = 2x = 30(2) = 60 ft

Check:

180 = 2(l + w)

180 = 2(60 + 30)

180 = 2(90)

180 = 180

LS = RS

4 0

3 years ago

If i have 2 c's 2 A's 3 b's and 1 d what is my gpa and would i be able to pass 8th grade

IgorC [24]

A= 4 points
B= 3 points
C= 2 points
D= 1 point
F= 0 points

So you add 2+2+4+4+3+3+3+1
Which equals 22
Now divide 22 by 8
You get 2.75
So yes with a 2.75 gpa you should pass 8th grade as most schools require nothing less than a 2.0

8 0

3 years ago