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anzhelika [568]
3 years ago
8

Please Help!!! Compare continuous functions f, g, and h, and match the statements with the function they best describe.

Mathematics
1 answer:
DiKsa [7]3 years ago
8 0

Answer: Not sure if this is 100% correct, but it should be.

Step-by-step explanation:

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Find the volume of the following solid figure. Use = 3.14. V = 4/3r3. A sphere has a radius of 3.5 inches. Volume (to the neares
Juli2301 [7.4K]
To solve the problem shown above, you must apply the proccedure shown below:

 1. You must use tthe formula for calculate the volume of a sphere, which is:

 V=4πr³/3

 V is the volume of the sphere.
 r is the radius of the sphere (r=3.5 inches)

 2. When you susbstitute these values into the formula shown above, you obtain the volume of the sphere. Therefore, you have:

 V=4πr³/3
 V=4π(3.5 inches)³/3

 3. Therefore, the answer is:

 V=179.5 inches³
4 0
3 years ago
Read 2 more answers
Please need help on this still lost
Korvikt [17]

Answer:

w = 30

Step-by-step explanation:

50 = 10 * 5 therefore w is 6 * 5 = 30

6/10 = 30/50

4 0
4 years ago
Read 2 more answers
Factor by grouping<img src="https://tex.z-dn.net/?f=%284y%5E3%20%2B%2028y%5E2%29%20%2B%20%28y%20%2B%207%29" id="TexFormula1" tit
Inessa05 [86]

We have the following expression given:

(4y^3+28y^2)+(y+7)

We can start selecting as common factor 4y^2 and we got:

4y^2(y+7)+(y+7)

Now we can select y+7 as common factor and we got:

(y+7)\left\lbrack 4y^2+1\right\rbrack

So then our final answer would be:

(y+7)(4y^2+1)

3 0
1 year ago
A grocery store sells jelly beans in the bulk section for $2.75 a pound. If Wanda buys 1.8 pounds, how much will it cost?​
mestny [16]

Answer:

$4.95

Step-by-step explanation:

If jelly beans cost $2.75 per pound then multiplying the cost per pound by the amount of jelly beans purchased will give the answer of $4.95 per 1.8 pounds of jelly beans.

5 0
4 years ago
Which proportion would you use to solve this problem?
erastovalidia [21]

Answer: 225

Step-by-step explanation:

225x0.04 is 9

4 0
3 years ago
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