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Sunny_sXe [5.5K]

3 years ago

15

There are 32 times students can work in the computer lab during one week. If 24 students can work in the computer lab at one tim

e, about how many students can work in the computer lab during one week?

2 answers:

Mrac [35]3 years ago

7 0

768 students can work in the computer lab during one week

uysha [10]3 years ago

4 0

600 students can work in the lab

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A car has a windshield wiper on the driver's side that has total arm length of 10 inches. It rotates

son4ous [18]

Answer:

75.44 Square Inches

Step-by-step explanation:

The diagram of the problem is produced and attached.

To determine the area of the cleaned sector:

Let the radius of the larger sector be R

Let the radius of the smaller sector be r

Area of the larger sector $=\frac{\theta}{360}X\pi R^2$

Area of the smaller sector $=\frac{\theta}{360}X\pi r^2$

Area of shaded part =Area of the larger sector-Area of the smaller sector

$=\frac{\theta}{360}X\pi R^2-\frac{\theta}{360}X\pi r^2\\=\frac{\theta \pi}{360}X (R^2- r^2)$

From the diagram, R=10 Inch, r=10-7=3 Inch, $\theta=95^\circ$

Therefore, Area of the sector cleaned

$=\frac{95 \pi}{360}X (10^2- 3^2)\\=75.44$ Square Inches$

7 0

3 years ago

G is proportional to the square of t. If t=2 and g=64, find g when t=3.5.

choli [55]

$If\ g\ is\ proportional\ to\ the\ t^2\ we\ have\ proportion\\\\
g=64\ \ \ \ ---\ \ \ t^2=2^2=4\\\\\
g=?\ \ \ ----t^2=(3,5)^ 2=12,25\\\\
4g=64*12,25\\\\
4g=784\ \ |:4\\\\g=196\\\\ Answer\ is\ g=196\ for\ t=3,5.$

4 0

3 years ago

A full moon occurs about every 30 days . If the last full moon occurred on a Friday, how many days will pass before a full moon

pentagon [3]

The answer is 30 days until it appears on a Friday again. In order to see the full moon again, we have to go through the entire cycle. So, it will be about 30 days before we would have seen another one.

7 0

3 years ago

Find the exact value of the trigonometric function given that sinU=-7/25 and cosV=-4/5 (Both U and V are in Quadrant III)

Contact [7]

cosU:-

$\\ \rm\longmapsto \sqrt{1-sin^2U}=\sqrt{1-49/625}=24/25$

As U lies in Q3

cosU=-24/25

sinV

$\\ \rm\longmapsto \sqrt{1-cos^2V}=\sqrt{1-16/25}=3/4$

As V lies in Q3

sinV=-3/5

So

sin(V-U)=sinVcosU-cosVsinU=(-3/5)(-24/25)-(-4/5)(-7/25)=72/125-28/125=72-28/125=<u>4</u><u>4</u><u>/</u><u>1</u><u>2</u><u>5</u>
cos(U-V)=cosUcosV+sinUsinV=(-24/25)(-4/5)+(-7/25)(-3/5)=96/125+21/125=117/125

3 0

2 years ago

Solve this equation. 1+1 = ?

max2010maxim [7]

Answer:

2

Step-by-step explanation:

5 0

3 years ago