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3 years ago

What's the answer for 2 and 3

Mathematics

1 answer:

Lorico [155]3 years ago

5 0

To find the unit rate you do distance ÷ time
16 miles ÷ 2 hours = 18
To find the time you do distance ÷ unit rate
20 miles ÷ 18 = 1.1
To find the distance you do time • unit rate
3 hours • 18 = 54

2) 1.1, 54, 1.8, 90, 27
3) 48, 3, 30, 2, 1

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Ramon drives his car 150 miles in 3 hours. Find the unit rate.

erma4kov [3.2K]

Answer:

The answer is A

Step-by-step explanation:

A) Ramon drive 50 miles per hour

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4 years ago

Write an equation of a line with the given slope and y-intercept

Verizon [17]

Y=mx+b
m=slope
b=yint

given
m=2
b=4/5
easy

y=2x+4/5

A

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3 years ago

Is 3:4 equivalent to 12:24

Nuetrik [128]

No because 3 times 4 is 12 but 4 times four is 16 and TO get from 4 to 24 you multiply 6 but 6 times 3 is 18 so no I think

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4 years ago

Read 2 more answers

Jenna can clean the gutters at her house in four hours. Her older brother, John, can clean the gutters in three hours, if they w

OlgaM077 [116]

Answer:

1 hour and 42.8 minutes

Step-by-step explanation:

To answer this question let's call $t_1$ while it takes Jenna to clean the gutters

Let's call $t_2$ while it takes John to clean the gutters

$t_1 = 4$ h

$t_2 = 3$ h

t = total time

g = job = 1 (clean the gutters)

The speed of each one is:

$V_1 = \frac{1}{4}$ g/ h

$V_2 = \frac{1}{3}$ g/h

$V = V_1 + V_2 = \frac{g}{t} = \frac{1}{t}$

So:

$V = V_1 + V_2 = \frac{1}{4} +\frac{1}{3}$

$\frac{1}{t} = \frac{1}{4} + \frac{1}{3}= \frac{7}{12} g/h$

$t = \frac{12}{7}$ h

Then, both together paint $\frac{7}{12}$ of gutters for each hour.

This means that it takes $\frac{12}{7}$ hours to clean the gutters together

Finally cleaning together takes 1,714 hours or also

1 hour and 42.8 minutes

7 0

3 years ago

Find the length of the curve y = integral from 1 to x of sqrt(t^3-1)

Arlecino [84]

$y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt$

By the fundamental theorem of calculus,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}$

Now the arc length over an arbitrary interval $(a,b)$ is

$\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx$

But before we compute the integral, first we need to make sure the integrand exists over it. $x^{3/2}$ is undefined if $x$ , so we assume $a\ge0$ and for convenience that $a$ . Then

$\displaystyle\int_a^bx^{3/2}\,\mathrm dx=\frac25x^{5/2}\bigg|_{x=a}^{x=b}=\frac25\left(b^{5/2}-a^{5/2}\right)$

6 0

3 years ago