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stira [4]
3 years ago
5

Can somebody help me please. It is a easy one

Mathematics
2 answers:
mylen [45]3 years ago
6 0

Answer:17

Step-by-step explanation:

Debora [2.8K]3 years ago
6 0
The answer to that question is 17.
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A math class has 3 girls and 7 boys in the 7th grade and 5 girls and 5 boys in the 8th grade the teacher randomly selects a seve
Minchanka [31]
In the 7th and 8th grade combined, there 3+5 girls = 8 girls, and 7+5 boys = 12 boys. If there are only boys and girls, then there are 12+8=20 students in all. 

The fraction of girls out of the total is P = 8/20 = 2/5.
8 0
2 years ago
Please help me out!!!!
Kobotan [32]

Answer:

All real numbers

Step-by-step explanation:

4(x+1) = 4x+4

Distribute

4x+4 = 4x+4

Subtract 4x from each side

4x+4-4x = 4x+4-4x

4=4

This is always true so there are infinite solutions

3 0
2 years ago
Read 2 more answers
Please help?
Mazyrski [523]

Answer:

23\sqrt{3}\ un^2

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\

2. Note that

A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0
3 years ago
PLEASE HELP!!!!! The velocity of sound in air is given by the equation , where v is the velocity in meters per second and t is t
jeyben [28]

The equation of velocity of sound in air is v = 20 √(273 + t). In this problem, we need to find the temperature when the velocity is 329 meters/s. You need to measure the time it takes a sound to travel a measured distance in order to measure its speed in air.

 

Given:

Velocity of sound in air equation = v = 20 √(273 + t)

Velocity = 329 m/s

 

To solve:

V = 20 √(273 + t)

<span>
329 = 20 * sqrt(273 + t)</span>

<span>
16.45 = sqrt(273 + t)</span>

<span>
273 + t = 16.45^2</span>

<span>
t- 16.45^2-273</span>

<span>
t = -2.4 degrees Celsius</span>

 

<span>So, the temperature when the velocity is 329 meters per second is -2.4 degrees Celsius.</span>

3 0
3 years ago
An individual head of sprinkling water system covers a circular area of grass with a radius of 25 ft the yard has three sprinkle
Greeley [361]

Answer:

Answer will be 5890 square feet. Step-by-step explanation: An individual head of a sprinkler system covers a circular area of grass with a radius of 25 feet. Now 3A = 3×π×625 feet²= 3×3.14×625 = 5887.5 ≈ 5890 square feet.

Step-by-step explanation:

Mark brainiest, thx

6 0
3 years ago
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