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agasfer [191]
3 years ago
14

30 is 2/3 of what number

Mathematics
2 answers:
alex41 [277]3 years ago
7 0

Answer:

Step-by-step explanation:

let the number be x.

then,

2/3 of x = 30

=> 2/3x = 30

=> x = 30 × 3/2

=> x = 45

so, the required number is 45 .

hope it helps...!!!

r-ruslan [8.4K]3 years ago
6 0

Answer:45

Step-by-step explanation:

Key Word:of

30*2/3 = 45

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11(11b + 2) what is the answer?
lisabon 2012 [21]

Answer:

121b+22

Step-by-step explanation:

Ok so first you distribute the 11 because its outside the ( ).

(11*11b) + (11*2)

121b+22

Since you don't know what b is you can't go any further, there for your done.

3 0
3 years ago
An employee has to pay 15.5% in taxes.If he earns $2,450 a week, how much will he have to pay in taxes
Neporo4naja [7]
$the employee will have to pay 379.75 in taxes
8 0
3 years ago
How do you expand (4x+2)2
kifflom [539]

Answer:

=8x+4

Step-by-step explanation:

=2(4x+2)

a=2,b=4x,c=2

=2 x 4x + 2 x 2

Simplify- 2 x 4x + 2 x 2: 8x + 4

6 0
3 years ago
PLZZ HELP!!!!!!
UNO [17]

Answer:

The answer is x = - 30, by using the <u>Subtraction Property of Equality</u>.

Step-by-step explanation:

Solve for x:

-2 = x + 28

Turn the equation over:

x + 28 = -2

-Subtract both sides by 28 by using <u>Subtraction Property of Equality</u>:

x + 28 - 28 = -2 - 28

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So, the final answer is x = -30 .

8 0
3 years ago
A balloon is blowing up at a constant rate of 9 cubic centimeters per second. When the volume of the balloon is 2048/3 pi cubic
jekas [21]

Answer:

\displaystyle \frac{dr}{dt}\approx 0,0112\ cm/sec

Step-by-step explanation:

<u>Rates of Change as Derivatives</u>

If some variable V is a function of another variable r, we can compute the rate of change of one with respect to the other as the first derivative of V, or

\displaystyle V'=\frac{dV}{dr}

The volume of a sphere of radius r is

\displaystyle V=\frac{4}{3}\pi r^3

The volume of the balloon is growing at a rate of 9\ cm^3/sec. This can be written as

\displaystyle \frac{dV}{dt}=9

We need to compute the rate of change of the radius. Note that both the volume and the radius are functions of time, so we need to use the chain rule. Differentiating the volume with respect to t, we get

\displaystyle \frac{dV}{dt}=\displaystyle \frac{dV}{dr}\displaystyle \frac{dr}{dt}

\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}

solving for \displaystyle \frac{dr}{dt}

\displaystyle \frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2}

We need to find the value of r, which can be obtained by using the condition that in that exact time

\displaystyle V=\frac{2048}{3}\pi\ cm^3

\displaystyle \frac{2048}{3}\pi=\frac{4}{3}\pi r^3

Simplifying and isolating r

\displaystyle r^3=512

\displaystyle r=\sqrt[3]{512}=8\ cm

Replacing in the rate of change

\displaystyle \frac{dr}{dt}=\frac{9}{4\pi 8^2}

\displaystyle \frac{dr}{dt}=\frac{9}{256\pi }

\displaystyle \frac{dr}{dt}\approx 0,0112\ cm/sec

8 0
3 years ago
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