Please help!!! What is the area of the trapezoid?

The distance from Earth to Mars is 136,000,000 miles. A spaceship travels at 31,000 miles per

posledela

136,000,000/31,000= how many hours for the spaceship to reach mars
136,000,000/31,000/24= how many days for the spaceship to reach mars
136,000,000/31,000/24 is approx. 182.7 days or 183 days (rounded up b/c of "to the nearest day").

7 0

3 years ago

Read 2 more answers

Last year there were 120 students in choir. This year, 30% more students took choir. How many students are taking choir this yea

hjlf

Answer:

120 into a third and divide and the answer

Step-by-step explanation:

160

7 0

3 years ago

what is the product of 9.8 x 10 to the power of 3 and 8.8 x 10 to the power of 6 expressed in scientific notation

sergij07 [2.7K]

Answer:

9.8*10^3

= 9,800

8.8*10^6

= 8,800,000

4 0

2 years ago

A maritime flag is shown. What is the area of the shaded part of the flag? Explain

Rom4ik [11]

Answer:

$Area = 72in^2$

Step-by-step explanation:

Your question is incomplete without an attachment (See attachment)

Required

Determine the area of the shaded part

From the attachment;

Assume that the shaded portion is closed to the right;

Calculate the Area:

$Area_1 = Length * Width$

$Area_2 = 8in * (6in + 6in)$

$Area_2 = 8in * 12in$

$Area_1 = 96in^2$

Next;

Calculate the Area of the imaginary triangle (on the right)

$Area_2 = \frac{1}{2} * base * height$

$Area_2 = \frac{1}{2} * (6in + 6in) * 4in$

$Area_2 = \frac{1}{2} * 12in * 4in$

$Area_2 = \frac{1}{2} * 48in^2$

$Area_2 = 24in^2$

Lastly, calculate the Area of the Shaded Part

$Area = Area_1 - Area_2$

$Area = 96in^2 - 24in^2$

$Area = 72in^2$

Hence,

The area of the shaded part is 72in²

3 0

2 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

5.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

8.

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

1 year ago