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liberstina [14]
3 years ago
6

What expression is equivalent to the area of metal sheet required to make this square-shaped traffic sign?

Mathematics
2 answers:
vovangra [49]3 years ago
7 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below are the choices based on the question above:

<span>16x2 + 12x + 9

  16x2 + 24x + 9

  8x2 + 9

  8x2 + 6
</span>
The answer is the second one <span>16x^2 +24x+9</span>
AveGali [126]3 years ago
3 0

Answer: \text{Area of the square shaped traffic sign }=16x^2+9+24x

Step-by-step explanation:

Given: The side of the square shaped traffic sign = 4x+3

We know that the area of square is given by:-

Area= side^2

Therefore, the area of the side of the square shaped field is given by:-

Area=(4x+3)^2

We know that , (a+b)^2=a^2+b^2+2ab

Therefore,

(4x+3)^2=(4x)^2+(3)^2+2(4x)(3)\\=16x^2+9+24x

Hence, \text{Area of the square shaped traffic sign }=16x^2+9+24x

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There is an equal number of men, women and children at a party. How many persons could be at the party?
astra-53 [7]

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3, 6, 9, 12, and so on, basically any number that can be divided by 3

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2 years ago
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First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x
e-lub [12.9K]

Answer:

(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

\int x ln(5+x)dx

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

\int (U-5) ln(U)dU

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

\int (pq')=pq-\int qp'

so we must define p, q, p' and q':

p=ln U

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and now we plug these into the formula:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU

Which simplifies to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU

Which solves to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C

so we can substitute U back, so we get:

\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C

and now we can simplify:

\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C

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7 0
3 years ago
One day at football practice, Darrell, the kicker, punted the ball so that its height in feet above the ground was given by the
nika2105 [10]

When the football is 81 feet high, it will be √  3 / 8 seconds

<h3>Function</h3>

Function relates input to output. Therefore,

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