Ask question

Ask question

All categories

3 years ago

6

The side length of a square is represented by the expression 2x + 5. Which expression represents the difference between the area

of the square and the perimeter of the square?

1 answer:

musickatia [10]3 years ago

7 0

Answer:

$4x^2 + 22x + 15$

Step-by-step explanation:

The side length of a square is represented by the expression 2x + 5.

The area of a square is given as:

$A = a^2$

where a = length of side of the square

The area of the square is therefore:

$A = (2x + 5)^2\\\\A = (2x + 5)(2x + 5)\\\\A = 4x^2 + 20x + 25$

The perimeter of a square is given as:

$P = 4a$

The perimeter of the square is therefore:

$P = 4(2x + 5) \\\\P = 8x + 10$

The difference between the area of the square and the perimeter of the square is:

$4x^2 + 30x + 25 - (8x + 10)\\\\4x^2 + 30x + 25 - 8x - 10\\\\4x^2 + 30x - 8x + 25 -10\\\\4x^2 + 22x + 15$

The expression that represents the difference between the area and the perimeter of the square is:

$4x^2 + 22x + 15$

You might be interested in

(a)Use differentiation to find the minimum value of f(x)= 3x^2 - 5x + 10 and the value(s) of x for which this minimum occurs.

elena-14-01-66 [18.8K]

Hi there!

Begin by differentiating f(x) using the power rule:

dy/dx = nxⁿ⁻¹

Therefore:

f(x) = 3x² - 5x + 10

f'(x) = 6x - 5

Set this equation equal to 0 to find the x-intercept:

0 = 6x - 5

5 = 6x

x = 5/6, which is where the graph goes from NEGATIVE to POSITIVE, so there is a MINIMUM at this value.

3 0

3 years ago

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

3 0

3 years ago

If you are 600 m from a batter when you see him hit a ball, how long will you wait to hear the sound of the bat hitting the ball

Molodets [167]

Answer:

<em>The speed of sound at 20°C is 343.42 m/s.</em>

<em>You have to wait 1.75 seconds to hear the sound of the bat hitting the ball</em>

Step-by-step explanation:

<u>Speed of Sound</u>

The speed of sound is not constant with temperature. Generally speaking, the greater the temperature, the greater the speed of sound.

The approximate speed of sound in dry air at temperatures T near 0°C is calculated from:

$\displaystyle v_{\mathrm {s} }=(331.3+0.606\cdot T )~~~\mathrm {m/s}$

The air is at T=20°C, thus the speed of sound is:

$\displaystyle v_{\mathrm {s} }=(331.3+0.606\cdot 20 )~~~\mathrm {m/s}$

$v_{\mathrm {s} }=343.42\ m/s$

The speed of sound at 20°C is 343.42 m/s.

To calculate the time to hear the sound after the batter hits the ball, we use the formula of constant speed motion:

$\displaystyle v=\frac{d}{t}$

Where d is the distance and t is the time. Solving for t:

$\displaystyle t=\frac{d}{v}$

Substituting the values v=343.42 m/s and d=600 m:

$\displaystyle t=\frac{600}{343.42}$

t = 1.75 s

You have to wait 1.75 seconds to hear the sound of the bat hitting the ball

6 0

3 years ago

Find the area + perimeter of the following triangle

White raven [17]

Answer:

Perimeter = 14 $\sqrt{2}$

Area = $\sqrt{42}$

Step-by-step explanation:

4 0

2 years ago

A 10ft by 20ft rectangular swimming pool is surrounded by a walkway of uniform width. If the total area of the walkway is 216ft^

Dmitry [639]

check the picture below.

so, we know the dimensions of the pool, is a 20x10, so its area is simply 200 ft², and we know the walkway is 216 ft², so the whole thing, including pool and walkway is really 200 + 216 ft².

now, as you see in the picture, the dimensions for the combined area is 20+2x and 10+2x, since the walkway is "x" long, therefore,

$\bf \stackrel{length}{(20+2x)}\stackrel{width}{(10+2x)}=200+216\implies \stackrel{FOIL}{4x^2+60x+200}=200+216 \\\\\\ 4x^2+60x=216\implies \stackrel{dividing~by~4}{x^2+15x=54} \\\\\\ x^2+15x-54=0\implies (x+18)(x-3)=0\implies x= \begin{cases} -18\\ \boxed{3} \end{cases}$

notice, it cannot be -18, since is a positive length unit.

5 0

3 years ago

Read 2 more answers