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2 years ago

Jade can do 65 push ups in 5 minutes. At the same rate, how many push ups can you expect her to do in 7 minutes?

Mathematics

2 answers:

hram777 [196]2 years ago

3 0

Answer:

Step-by-step explanation:

Set up a proportion, where x is the number of push ups she can do in 7 minutes

$\frac{65}{5}$ = $\frac{x}{7}$

Cross multiply and solve for x:

5x = 455

x = 91

So, Jade can do 91 push ups in 7 minutes

musickatia [10]2 years ago

3 0

Answer:

She can do 91 pushups

Step-by-step explanation: make an equation

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Write and evaluate the definite integral that represents the volume of the solid formed by revolving the region about the x-axis

guapka [62]

Answer: V = $\frac{64}{3}\pi$

Step-by-step explanation: A solid formed by revolving the region about the x-axis can be considered to have a thin vertical strip with thickness Δx and height y = f(x). The strip creates a circular disk with volume:

V = $\pi. y^{2}.$ Δx

Using the <u>Disc</u> <u>Method</u>, it is possible to calculate all the volume of these strips, giving the volume of the revolved solid:

V = $\int\limits^a_b {\pi. y^{2} } \, dx$

Then, for the region generated by y = - x + 4:

V = $\int\limits^4_0 {\pi.(-x+4)^{2} } \, dx$

V = $\pi.\int\limits^4_0 {(x^{2}-8x+16)} \, dx$

V = $\pi.(\frac{x^{3}}{3}-4x^{2}+16x )$

V = $\pi.(\frac{4^{3}}{3}-4.4^{2}+16.4 - 0 )$

V = $\frac{64}{3}.\pi$

The volume of the revolved region is V = $\frac{64}{3}.\pi$

8 0

2 years ago

A particle moves on a straight line and has acceleration a(t)=24t+2. Its position at time t=0 is s(0)=3 and its velocity at time

user100 [1]

Answer:

It's position at time t = 5 is 593.

Step-by-step explanation:

The velocity v(t) is the integral of the acceleration a(t)

The position s(t) is the integral of the velocity v(t)

We have that:

The acceleration is:

$a(t) = 24t + 2$

Velocity:

$v(t) = \int {a(t)} \, dt = \int {24t + 2} \, dt = 12t^{2} + 2t + K$

K is the initial velocity, that is v(0). Since V(0) = 13, K = 13

Then

$v(t) = 12t^{2} + 2t + 13$

Position:

$s(t) = \int {s(t)} \, dt = \int {12t^{2} + 2t + 13} \, dt = 4t^{3} + t^{2} + 13t + K$

Since s(0) = 3

$s(t) = 4t^{3} + t^{2} + 13t + 3$

What is its position at time t=5?

This is s(5).

$s(t) = 4t^{3} + t^{2} + 13t + 3$

$s(5) = 4*5^{3} + 5^{2} + 13*5 + 3$

$s(5) = 593$

It's position at time t = 5 is 593.

3 0

2 years ago

Solve for x: -3 + x = -22

Tanya [424]

Answer:

x= -19

Step-by-step explanation:

-3 + x = -22

Add 3 to each side

-3+3 +x = -22 +3

x = -19

3 0

2 years ago

Can someone plz help me with this

slava [35]

Answer:

Step-by-step explanation:

2a. f(-3) = (-3)^2-4(-3)+6 = 9+12+6 = 27

2b. f(-2) = 3(-2)^2-(-2)^3 -2 +1 = 3(4) - (-8) -2 +1

= 12+8-1 = 19

8 0

2 years ago

Sarah jogged 4.8 miles each day for 22 days last month. How many miles did Sarah jog in total?

solong [7]

Answer:

105.6

hope this helps

have a good day :)

Step-by-step explanation:

8 0

2 years ago