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2 years ago

Which statement best describes (3 x 4 x6 = 3 x (4 x 6)

Mathematics

1 answer:

Ratling [72]2 years ago

7 0

Answer:

The given statement is,

(3 * 4) * 6 = 3(4 * 6)

To find (3 * 4) * 6

(3 * 4) * 6 = 12 * 6 = 72

To find 3(4 * 6)

3(4 * 6) = 3 * 24 = 72

Step-by-step explanation:

The result of these two expression is same.

Therefore we can conclude that,

Therefore the correct answer is,

The grouping of the factors is changed but the product is the same.

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Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

4 0

3 years ago

Simplify the expression. Explain each step.<br><br> 3.2+(x+5.1)

Andrew [12]

Ok so I’m not sure but here’s what I would do to solve it
There is only one x so
X+
5.1+3.2=8.3
So
X+8.3 is the answer
Hope that helps xx

5 0

3 years ago

Mr. Mills is 5 feet and 11 inches tall. How many inches tall is he?

MrRa [10]

Answer:

Step-by-step explanation:

There are 12 inches in a foot:

12 inches x 5 feet = 60

60 + 11 inches = 71 inches total

3 0

3 years ago

Read 2 more answers

If 6 is added to three quarters of a certain number, the result is 42 . Find the original number

Basile [38]

Answer:

Step-by-step explanation

Hello!

3x/4 + 6=42

3x/4=42-6 = 36

3x = 36*4

3x = 144

x= 144/3 =48

3/4 of 48 = (48*3) / 4 =36

36+6=42

the original number is 48

4 0

3 years ago

Read 2 more answers

In the numerical sentence below, A = 8, D = 1/2 , and F = 1/4. What are the values B, C, and E if the sentence below is equivale

a_sh-v [17]

B=3; C=16; and E=4096.

Using the information we have gives us

$8^B \times C^{\frac{1}{2}} = E^{\frac{1}{4}}$

We know that:
$8^B = 2;
\\
\\C^{\frac{1}{2}}=4;
\\
\\E^{\frac{1}{4}}=8$

We know that 2³ = 8, so B = 3.

To "undo" a square root, we square the number; 4²=16, so C = 16.

To "undo" a fourth root, we raise the answer to the fourth power; 8⁴ = 4096, so E = 4096.

3 0

3 years ago

Read 2 more answers