Answer this question please I need help

A mirror should be centered on a wall. The mirror is 4 feet wide and the wall is 20 feet wide. Which equation helps determine th

Alinara [238K]

A.) x+4+x=20 should be the answer

7 0

2 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

I really need help with this question

Sergeu [11.5K]

Answer:

39

Step-by-step explanation:

I assume the father's age is a 2-digit number. He is not 9 years old or younger, and he is not 100 years old or older.

The sum of the digits of a two-digit number goes up by 1 from one number to the next unless the number ends in 9.

For example, from 17 to 18, the sum of the digits goes from 8 to 9.

If the number ends in 9, for example, 29, the digits add to 11. Then the next number is 30, and now the sum of the digits is 3, which is less than 11.

The father's age now is a number whose sum of digits is not only greater than the sum of the digits next year, but it is 3 times greater.

Let's look at 2-digit number and the next number and the sum of their digits.

19, 20; sums: 10, 2; 10/2 = 5, not 3

29, 30; sums: 11, 3; 11/3 ≠ 3

39, 40; sums: 12, 4; 12/4 = 3

Answer: The father is 39 years old.

5 0

2 years ago

Read 2 more answers

IAS 36. Impairment of asset. we are a photo

romanna [79]

Answer:

where is the photo? sorry i want to help im just confused where.

3 0

3 years ago

Read 2 more answers

Find the diameter of a cone that has a volume of 104.67 cubic inches and a height of 4 inches. Use 3.14 for pi. 4 inches 6 inche

aleksandr82 [10.1K]

V: volume of a cone = (πr²h)/3 = 104.67 in³
π: pi = 3.14
r: radius = 1/2 diameter = [unknown]
h: height = 4 in

V = (πr²h)/3
V = r²(πh)/3

r² = (3V)/(πh)
r² = (3 ×104.67)/(3.14 × 4)
r² = 25

r = √25
r = 5 (but remember the radius is only 1/2 the diameter)

thus . . .

<u><em>d = 10 in </em></u>

8 0

3 years ago

Read 2 more answers