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3 years ago

Answer ASAP PLEASE WILL MARK BRAINLIEST

Mathematics

2 answers:

NARA [144]3 years ago

4 0

Answer:

Answer should be A

Step-by-step explanation:

-7/2 divided by 49/4

irina1246 [14]3 years ago

3 0

Change in temperature per hour would be found by dividing the change in temperature by the amount of time.

Rewrite the numbers as improper fractions and the answer would be d.

-49/4 / 7/2

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Water has a freezing point of 32 degrees. Murcury has a freezing point that is 70 degrees lower. What is the freezing point of m

Mekhanik [1.2K]

Answer:

-38 degrees

Step-by-step explanation:

32-70

= -38

8 0

3 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

Y-4x=9 in standard form

11111nata11111 [884]

Answer: y - 4x = 9 <em>in standard form is</em> -y + 4x = -9

4 0

3 years ago

The total bill for dinner, before tax, in $36.00. If you leave a 20% tip, will you will leave a tile of $_____.

qaws [65]

Answer:

After the tip you would leave $43.20

Step-by-step explanation:

In order to find this, we must first find 20% of 36. In order to do that, you must multiply them together (don't forget to turn the percentage into a decimal first).

36 * .20 = 7.20

Now we can add the total amount pre-tip and the tip amount to find the overall total.

$36 + $7.20 = $43.20

8 0

3 years ago

How much miles is a light year

skelet666 [1.2K]

1 light-year = roughly 5,874,601,700,000 miles. (rounded)

To work it out, take the speed of light, and change units a few times:

      (186,282.397 miles/second)
times
          (86,400 second/day)
times
          (365 day/year)

=   (186,282.397 x 86,400 x 365) mile/year .

You can whip that out on your calculator in no time.

8 0

3 years ago