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romanna [79]
3 years ago
6

HELP!!!!!!!!!! What is the product? −5⋅(−3)⋅(−8) a−120 b−16 c16 d120

Mathematics
2 answers:
marta [7]3 years ago
8 0
A) -120 is the answer
Greeley [361]3 years ago
4 0

The answer is

a−120

because -5*-3=15 because negative and a negative = a positive

Now 15*-8 =-120 that's because a negative and a positive

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Find the 2 different rates at which the farm worker worked:

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Then you need to subtract 0.67 pint/min from 0.75 pint/min to answer this question.

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3 years ago
What is x given angle ABC~ angle DBE~ ?
Dovator [93]

30:50 = 3:5

x : x+25 = 3:5

25/2 x 3 = 37.5

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Safara wants to rent a bicycle the graph shows the cost for renting a bicycle for different lengths of time. What is the slope o
enot [183]

The slope of the line will be equal to 30.

Slope of a line may be defined as the steepness of a curve. Slope of line can be positive, negative or zero. Slope of a line is used to determine the equation of the line which is given as y = mx + b where m is the slope, b is y intercept and x and y are independent and dependent variables respectively. The slope of a line on points (x₁, y₁) and (x₂, y₂) is given by the formula, m = y₂ - y₁/x₂ - x₁.

According to the question the value of points is (x₁, y₁) = (0, 0) and (x₂, y₂) = (2, 60).

Now, the slope can be calculated as

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brainly.com/question/6531433

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4 0
1 year 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}}

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-Hope this helped! :D
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