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3 years ago

Picture Question please help!!!! Thank you so much!

Mathematics

2 answers:

PIT_PIT [208]3 years ago

7 0

33 because 33 times 3 is 99 however it can't be 100 wince there is no multiplication that can be a hundred by timsing the 3

ra1l [238]3 years ago

6 0

The answer is 33. This is because there are 3 colours and the probability that the spinner will land on each colour is 1/3 (one third). To work out the number of times it will land on blue you have to do the probability of it landing on blue (1/3) multiplied by the number of times spun (100) which gives approximately 33.

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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

3 years ago

riel wants to use squares to tile a floor that measures 240 cm by 300 cm what is the length of the largest square he can use

Alexxx [7]

So, what you want is the biggest square that can evenly tile a space 240*300. The biggest possible number that can divide 240 and 300. Does that ring any bells? Maybe you should divide 240 and 300

4 0

3 years ago

Differentiate the function with respect to x f(x)= x^4/ x^2- 4 Show work

jeyben [28]

Answer:

Step-by-step explanation:

$f(x)=\frac{u}{v} ,u~ and~ v~ functions~ of~ x\\f'(x)=\frac{v ~u'-u~v' }{v^2} \\f(x)=\frac{x^4}{x^2-4} \\f'(x)=\frac{(x^2-4)*4x^3-x^4(2x)}{(x^2-4)^2} \\=\frac{2x^3(2x^2-8-x^2) }{(x^2-4)^2} \\=\frac{2x^3(x^2-8)}{(x^2-4)^2}$