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2 years ago

A stray dog ate 6 of your muffins. That was four thirds of all of them. With how many did you you start?

Mathematics

1 answer:

Crazy boy [7]2 years ago

4 0

Answer:

Step-by-step explanation:

The total number of muffins was

Explanation:

You can translate the important parts of the sentence from English to math (say that

is the total number of muffins):

A stray dog ate



6o f your muffins.



That was





all of them!

Now copy down the equation and solve for

120

This means that the total number of muffins was

(before the stray dog ate them).

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Select all the values of x so g(x) is not a function.

alexdok [17]

The answer is 6 and 9
Step-by-step explanation: We are trying to figure out what makes the chart not a function. A function is is one input equals one output.
so all you have to do is look at the x and look at the number that is already there and just make it repeat.

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3 years ago

If cos x = sin(20 + x) ° and 0°

Varvara68 [4.7K]

Answer:

x = 35°

Step-by-step explanation:

The question is as following

cos x = sin(20 + x)° and 0° < x < 90° , find X?

==============================================

cos x = sin(20 + x)°

sin and cos are co-functions,

which means that: cos x = cos [90 - (20 + x)]

∴ x = 70 - x

∴ 2x = 70

∴ x = 35°

======================

Note: cos θ = sin ( 90 - θ )

4 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,

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