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2 years ago

Simplify 2x2a^2 x2a^2

Mathematics

1 answer:

horrorfan [7]2 years ago

6 0

Step-by-step explanation:

> 2×2a²×2a²

8a⁴

> 36a³×1/4a²

9a³×1/a²

> 2⁶

> 5²m²

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Over which interval is the graph of f(x) = one-halfx2 + 5x + 6 increasing?

kaheart [24]

Answer:

$\large\boxed{x\in[-5,\ \infty)}$

Step-by-step explanation:

$f(x)=\dfrac{1}{2}x^2+5x+6\\\\\text{The coeficient of}\ x^2\ \text{is the positive number. Therefore the parabola is op}\text{en up}.\\\\\text{If a parabola op}\text{en up, then the graph increasing in interval}\ (h,\ \infty),\\\text{and decreasing in interval}\ (-\infty,\ h).\ \text{Where}\ h\ \text{is a first coordinate of a vertex.}\\\\\text{For}\ y=ax^2+bx+c,\ h=\dfrac{-b}{2a}.\\\\\text{We have}\ a=\dfrac{1}{2}\ \text{and}\ b=5.\ \text{Substitute:}\\\\h=\dfrac{-5}{2\left(\frac{1}{2}\right)}=-\cdot\dfrac{5}{1}=-5.$

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3 years ago

Read 2 more answers

Find the quotient -7/12 divided by 1/7

leva [86]

Answer:

-4.08

Step-by-step explanation:

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2 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

2 years ago

Three numbers add up to 180. The second number is twice the first, and the third number is three times the first. WHAT IS EACH N

Ratling [72]

The answer is 30,60,90. Because 60 is twice the number 30 and 90 is three times the first number.

8 0

3 years ago

Use differentials to estimate the amount of metal in a closed cylindrical can that is 26 cm high and 10 cm in diameter if the me

Afina-wow [57]

Answer:

The estimated amount of metal in the can is 87.96 cubic cm

Step-by-step explanation:

We can find the differential of volume from the volume of a cylinder equation given by

$V= \pi r^2 h$

Thus that way we will find the amount of metal that makes up the can.

Finding the differential.

A small change in volume is given by:

$dV =\cfrac{\partial V}{\partial h} dh + \cfrac{\partial V}{\partial r} dr$

So finding the partial derivatives we get

$dV =\pi r^2 dh + \pi 2r h dr$

$dV =\pi r^2 dh + 2\pi r h dr$

Evaluating the differential at the given information.

The height of the can is h = 26 cm, the diameter is 10 cm, which means the radius is half of it, that is r = 5 cm.

On the other hand the thickness of the side is 0.05 cm that represents dr = 0.05 cm, and the thickness on both top and bottom is 0.3 cm, thus dh = 0.3 cm +0.3 cm which give us 0.6 cm.

Replacing all those values on the differential we get

$dV =\pi 5^2 (0.6) + 2\pi (5) (26) (0.05)$

That give us

$V= 28 \pi \, cm^3$

Or in decimal value

$\boxed{dV= 87.96 \, cm^3}$

Thus the volume of metal in the can is 87.96 cubic cm.

6 0

3 years ago