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3 years ago

If 9:x = x:4, then x is 6

Mathematics

1 answer:

maria [59]3 years ago

7 0

Answer:

$x = \sqrt{54}$

Step-by-step explanation:

$\frac{9}{x} = \frac{x}{6} \implies x^{2} = 9 \times 6 \implies x^{2} = 54 \implies x = \sqrt{54} \cong 7.35$

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Find the distance between -4,2 and -4,-5

My name is Ann [436]

Answer:

I think the distance is 7

7 0

2 years ago

In circle A shown below, the measure of ∠BAD is 148°: Circle A with angle BAD measuring 148 degrees; points B, C, D lie on Circl

Semmy [17]

Answer:

$m\ arc\ CD=43\°$

Step-by-step explanation:

we know that

$m$ -----> by central angle

see the attached figure to better understand the problem

we have

$m\ arc\ BC=105\°$

$m$

substitute

$148\°=105\°+m\ arc\ CD$

$m\ arc\ CD=148\°-105\°=43\°$

4 0

3 years ago

Read 2 more answers

Yesterday, the movie theater brought in $1,440 in ticket sales and $295.25 in food sales. Today, ticket sales were three-fourths

Mice21 [21]

Answer:

Nina is not correct because 393.5$ less then yesterday's ticket sales.

Step-by-step explanation:

To find total sales she he is not correct because

3 / 4 of 1440 = 1,080$ ticket sales from yesterday

and 295.25 - 33.50 = 261.75$

Therefore, ( 3 / 4 * 1440 ) + ( 295.25 - 33.50 ) = 1,341.75$ sale of today

Ticket from yesterday and today

1,080 - 1,341.75 = 393.5

393.5$ less then yesterday's ticket.

6 0

2 years ago

Read 2 more answers

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

Please help me with my math homework

Hatshy [7]

Answer:

So basically the mean is the average number.

Step-by-step explanation:

add all of them then divide by 6 (aka how many fractions)

8 0

2 years ago