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3 years ago

Evaluate the expression 5(7 − 4)2 ÷ 3 + 11.

Mathematics

2 answers:

Debora [2.8K]3 years ago

6 0

Answer:

Step-by-step explanation:

Parenthesis first: 5(3)(2)/3

15(2)/3

30/3

=10

Leviafan [203]3 years ago

3 0

The expression 5(7 − 4)2 ÷ 3 + 11 is equal to 20.9

<u>Solution:</u>

Given expression is 5(7 − 4)2 ÷ 3 + 11

We need to evaluate the given expression

Let us use "BODMAS" rule

BODMAS means sequence of operations which are Brackets, Order, Division, Multiplication, Addition and then Subtraction.

Now solve the given expression

5(7 − 4)2 ÷ 3 + 11

First "brackets" are to be solved as per BODMAS

5(3)2 ÷ 3 + 11

Now division is performed, 2 is divided by 3 to give 0.66

$=5(3) \times 0.66+11$

Now "brackets" is solved

$15 \times 0.66 + 11$

Now multiplication is performed

9.9 + 11

Atlast, addition is performed

20.9

Hence 5(7 − 4)2 ÷ 3 + 11 is equal to 20.9

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3 years ago

Calculus 2 Master needed, evaluate the indefinite integral of: <img src="https://tex.z-dn.net/?f=%5Cint%5C%28%20%28lnx%29%5E2%7D

viva [34]

Answer:

$\int (\ln(x))^2dx=x(\ln(x)^2-2\ln(x)+2)+C$

Step-by-step explanation:

So we have the indefinite integral:

$\int (\ln(x))^2dx$

This is the same thing as:

$=\int 1\cdot (\ln(x))^2dx$

So, let's do integration by parts.

Let u be (ln(x))². And let dv be (1)dx. Therefore:

$u=(\ln(x))^2\\\text{Find du. Use the chain rule.}\\\frac{du}{dx}=2(\ln(x))\cdot\frac{1}{x}$

Simplify:

$du=\frac{2\ln(x)}{x}dx$

And:

$dv=(1)dx\\v=x$

Therefore:

$\int (\ln(x))^2dx=x\ln(x)^2-\int(x)(\frac{2\ln(x)}{x})dx$

The x cancel:

$=x\ln(x)^2-\int2\ln(x)dx$

Move the 2 to the front:

$=x\ln(x)^2-2\int\ln(x)dx$

(I'm not exactly sure how you got what you got. Perhaps you differentiated incorrectly?)

Now, let's use integrations by parts again for the integral. Similarly, let's put a 1 in front:

$=x\ln(x)^2-2\int 1\cdot\ln(x)dx$

Let u be ln(x) and let dv be (1)dx. Thus:

$u=\ln(x)\\du=\frac{1}{x}dx$

And:

$dv=(1)dx\\v=x$

So:

$=x\ln(x)^2-2(x\ln(x)-\int (x)\frac{1}{x}dx)$

Simplify the integral:

$=x\ln(x)^2-2(x\ln(x)-\int (1)dx)$

Evaluate:

$=x\ln(x)^2-2(x\ln(x)-x)$

Now, we just have to simplify :)

Distribute the -2:

$=x\ln(x)^2-2x\ln(x)+2x$

And if preferred, we can factor out a x:

$=x(\ln(x)^2-2\ln(x)+2)$

And, of course, don't forget about the constant of integration!

$=x(\ln(x)^2-2\ln(x)+2)+C$

And we are done :)

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3 years ago