Which statement is not true

Choose the ratio equivalent to 10/9

satela [25.4K]

<h3>Answer:</h3>

30/27, 20/18...

<h3>Solution:</h3>

There are <em>infinitely many ratios equivalent to 10/9.</em>
Here are <em>some </em>of them:
30/27
20/18
100/90
60/54

Hope it helps.

Do comment if you have any query.

3 0

2 years ago

What is the value of 8 in the number 33.086?

Paha777 [63]

Answer:

The value of 8 in the number 33.086 would be the hundredth

Step-by-step explanation:

30 would be the ten

3 would be the one

.0 would be tenth

.8 is the hundredth

.6 is the thousandth

3 0

3 years ago

Read 2 more answers

How are circumference and area the same? How are they different?

Sindrei [870]

In geometry, the circumference (from Latin circumferences, meaning "carrying around") of a circle is the (linear) distance around it. That is, the circumference would be the length of the circle The area of a circle is pi (approximately 3.14) times the radius of the circle squared.

3 0

3 years ago

1- The acceleration of a particle is defined by the relation a = 6 ft/s 2. Knowing

TiliK225 [7]

Answer:

The initial velocity of the particle is

v' = -6 - 6.2 = -18 ft/s

at t = 5s, the velocity is v = -18 + 6.5 = 12 ft/s

the position is

$x = - 32 - 18 \times 5 + \frac{1}{2} \times 6 \times 5 {}^{2} = - 47 \: ft$

the total distance traveled is

$s = | \frac{12 {}^{2} - 18 {}^{2} }{2 \times 6} | = 15 \: ft$

8 0

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 $[0,1) \cup (3,5)$

5 0

3 years ago