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

2 answers:

7 0

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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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Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .