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2 years ago

2. Which ratio is equivalent to 4:9?

Mathematics

2 answers:

vlada-n [284]2 years ago

7 0

Answer:

Below!

Step-by-step explanation:

To find out which ratio is equal to 4:9, multiply anything to both the numerator and the denominator. Remember, the multiplication must be equal to both the numerator and the denominator must be the same. Please check below to understand this problem better.

=> 4:9 = 4/9 = 4 x 2/9 x 2 = 8/18 = 8:18
=> 4:9 = 4/9 = 4 x 3/9 x 3 = 12/27 = 12:27

These are all examples of ratios equivalent to 4:9

Hoped this helped!

cestrela7 [59]2 years ago

5 0

Answer:

4 : 9 is an equivalent ratio of 8 : 18.

Step-by-step explanation:

hope this helps

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2 years ago

(2/5)^-2(-3)^4 What is the equivalent?

Nikolay [14]

Answer:

-8100

Step-by-step explanation:

(2/5)^-2(-3)^4

Given data

and second term

The first term= (2/5)^-2 and

Second term=(-3)^4

Simplify the first term

(2/5)^-2= 2^-2/ 5^-2= 1/2^2 * 1/5^2

=1/4/ 1/25

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2 years ago

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

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3 years ago

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Answer: J=12

Explanation: To make this equation easier to solve, you can make everything a whole number (a.k.a. Remove the fractions). To do so, you can multiply everything in this equation by the Least Common Multiple (LCM) of all the denominators, which is 12. The equation now becomes:

-4=3J-40

Then, solve it using normal algebra:

-4=3J-40
36=3J
J=12

Much easier, right?!

Hope that helps シ

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2 years ago