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3 years ago

Q4: John runs 15 miles in 3 hours. How many miles can John run per hour?

Mathematics

2 answers:

denis-greek [22]3 years ago

8 0

Answer:

5 per hour

Step-by-step explanation:

15 miles the total amount he ran divided by 3 the total amount of time

15/3=5 so John ran 5 mph

spin [16.1K]3 years ago

8 0

Answer:

john runs 5 miles per hour because 3*5=15

Step-by-step explanation:

You might be interested in

Solve the Anti derivative.

Alex Ar [27]

Answer:

$\displaystyle \int {\frac{1}{9x^2+4}} \, dx = \frac{1}{6}arctan(\frac{3x}{2}) + C$

General Formulas and Concepts:

Algebra I

Factoring

Calculus

Antiderivatives - integrals/Integration

Integration Constant C

U-Substitution

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Trig Integration: $\displaystyle \int {\frac{du}{a^2 + u^2}} = \frac{1}{a}arctan(\frac{u}{a}) + C$

Step-by-step explanation:

Step 1: Define

$\displaystyle \int {\frac{1}{9x^2 + 4}} \, dx$

Step 2: Integrate Pt. 1

[Integral] Factor fraction denominator: $\displaystyle \int {\frac{1}{9(x^2 + \frac{4}{9})}} \, dx$
[Integral] Integration Property - Multiplied Constant: $\displaystyle \frac{1}{9} \int {\frac{1}{x^2 + \frac{4}{9}}} \, dx$

Step 3: Identify Variables

Set up u-substitution for the arctan trig integration.

$\displaystyle u = x \\ a = \frac{2}{3} \\ du = dx$

Step 4: Integrate Pt. 2

[Integral] Substitute u-du: $\displaystyle \frac{1}{9} \int {\frac{1}{u^2 + (\frac{2}{3})^2} \, du$
[Integral] Trig Integration: $\displaystyle \frac{1}{9}[\frac{1}{\frac{2}{3}}arctan(\frac{u}{\frac{2}{3}})] + C$
[Integral] Simplify: $\displaystyle \frac{1}{9}[\frac{3}{2}arctan(\frac{3u}{2})] + C$
[integral] Multiply: $\displaystyle \frac{1}{6}arctan(\frac{3u}{2}) + C$
[Integral] Back-Substitute: $\displaystyle \frac{1}{6}arctan(\frac{3x}{2}) + C$

Topic: AP Calculus AB

Unit: Integrals - Arctrig

Book: College Calculus 10e

7 0

2 years ago

Given that(4,6) is on the graph of f(x) find the corresponding point for the function f(-4x)

Len [333]

Answer:

(-1,6)

Step-by-step explanation:

Given that(4,6) is on the graph of f(x)

f(-4x) means x is multiplied by -4

When x is multiplied by -1 then there will be reflection over y axis

We multiply every point by -1. so multiply the x values of the given point (4,6) by -1

New point is (-4,6)

If any number is multiplied with x then there will be a horizontal compression or stretch.

4 is multiplied with x , so there will be horizontal compression because 4 is greater than 1

To get new point, we divide the x values by 4 for compression

we already got (-4,6) after multiplying by -1

Now we divide the x coordinate -4 by 4 = -1

So corresponding point for the function f(-4x) is (-1,6)

5 0

3 years ago

I’ll give brainliest, just help me please :)

Lemur [1.5K]

Answer:

V = 408 SA = 378

Step-by-step explanation:

To find the volume, you need to first find out the area and multiply it by overall length.

A = 1/2 (6)(8)

= 24

Volume = 24 x 17

= 408

Surface Area

SA = Front and Back + Right Side + Left Side + Bottom

= 2 [1/2 (6) (8)] + (17 x 10) + (3 x 8) + (17 x 8)

= 2 (24) + 170 + 24 + 136

= 48 + 170 + 24 + 136

= 378

6 0

2 years ago

Find the area of the shaded regions. Give your answer as a completely simplified exact value in terms of π (no approximations).

Elan Coil [88]

Answer:

(40/3)*pi

Step-by-step explanation:

lets first find the area of the little circle

A=pi*r^2

A=pi*3^2

A=9pi

Now, lets find the area of the big circle

A=pi*r^2

A=pi*(3+4)^2

A=pi*7^2

A=49pi

now lets subtract the area of the little circle from the area of the big circle

49pi-9pi=40pi, now we found the area of the "bagel)

lets find what portion of the bagel is the shaded region

120/360=1/3

now lets multiply the bagel area by the fraction

40pi*(1/3)=

(40/3)*pi or (40*pi)/3

4 0

3 years ago

Please help. What is the circumference of the circle? (X+4)^2+(y-5)^2=81

PIT_PIT [208]

the radius: r^2 = 81

r =9

c = 2*pi*r

= 2 * pi * 9

= 18 *pi

Answer: 18*pi or approximately 56.62

8 0

3 years ago

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