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2 years ago

How do you do this thanks

Mathematics

1 answer:

elixir [45]2 years ago

3 0

Answer:

hmmm i might know this i think number 1 is the right answer im guessing (o_o)

Step-by-step explanation:

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a baseball pitcher won 80% of the games he played. if he pitched 35 games, how many games did he win?

andrey2020 [161]

Answer:

28 games

Step-by-step explanation:

5 0

2 years ago

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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

2 years ago

If you add one-third of a number to the number itself, you get 16.what is the number?

GuDViN [60]

The answer to this is 12.

One third of 12 is 4.

12+4=16

Comment if you have ANY questions!

Hope this helps!

7 0

2 years ago

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Use the method of completing the square to solve x2 - 6x + 1 = 0. Simplify your solutions.

exis [7]

Answer:

x = 3 ± 2 $\sqrt{2}$

Step-by-step explanation:

given

x² - 6x + 1 = 0 ( subtract 1 from both sides )

x² - 6x = - 1

To complete the square

add (half the coefficient of the x- term)² to both sides

x² - 6x + (- 3)² = - 1 + (- 3)²

x² - 6x + 9 = - 1 + 9

(x - 3)² = 8 ( take the square root of both sides )

x - 3 = ± $\sqrt{8}$ = ± 2 $\sqrt{2}$ ( add 3 to both sides )

x = 3 ± 2 $\sqrt{2}$

8 0

2 years ago

HURRY!!!!!!! Which symbol replaces ? to make the statement true? 16÷4+12 ? 4+24÷2 A.< B.= C.>

satela [25.4K]

16 / 4 + 12 ___ 4 + 24/2
4 + 12 ___ 4 + 12
16 __ 16

u need an equal sign because 16 = 16

3 0

2 years ago

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