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3 years ago

6

Ally's hair grew from 10 10 \frac{3}{4} 4 3 inches to 13 13 \frac{1}{4} 4 1 inches over 5 months. At what rate did Ally's ha

ir grow per month?

1 answer:

Rzqust [24]3 years ago

3 0

0.5 inches a month.

I believe you subtract 13 1/4 by 10 3/4 = 2 1/2.

Then you divide by 5.

You might be interested in

A survey of middle school students asked if they play video games. The responses are

kow [346]

Answer:

A. 35.69%

B. 62.77%

Step-by-step explanation:

In this case, what we must do is calculate the probability supported with the data in the table.

Part A.

What is the probability that a randomly selected student is a 7th grader?

Here the total is all the students that would be the sum of all the grades, which are 325 students.

the number of 7th graders is 116, so the probability is:

116/325 = 0.3569

that is, 35.69%

Part B: What percentage of the students interviewed play Video games?

The total number of students is the same 325 and of those 204 play video games, therefore:

204/325 = 0.6277

that is, 62.77%

6 0

3 years ago

Factor -8x^3-2x^2-12x-3

inysia [295]

$-8x^3-2x^2-12x-3\\\\=-2x^2(4x+1)-3(4x+1)\\\\=(4x+1)(-2x^2-3)\\\\=\boxed{-(4x+1)(2x^2+3)}$

5 0

3 years ago

Please see attachment

Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

<u>Step(i)</u>:-

Given function

$f(x) = \frac{-x}{2x^{2} +1}$ ...(i)

Differentiating equation (i) with respective to 'x'

$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$2 x^{2}-1 = 0$

$2 x^{2} = 1$

$x^{2} = \frac{1}{2}$

$x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

<u><em>Step(ii):</em></u>-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

<u><em>Step(iii):</em></u>-

The value of absolute minimum value

$f(x) = \frac{-x}{2x^{2} +1}$

$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

<u><em>Final answer</em></u>:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

3 0

2 years ago

Could you please help me I don’t really understand this pleaseeeeeee?

likoan [24]

B has to be 70° and A has to be 115° so I’m confused by your options

7 0

3 years ago

Find the volume of the rectangular prism. WA 3yd 3 yd 4 yd

dlinn [17]

Answer:

V = 36

Step-by-step explanation:

V=whl=3·3·4=36

6 0

2 years ago

Read 2 more answers