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Volgvan
3 years ago
7

72 1/10% of 19 =Show work

Mathematics
1 answer:
Lerok [7]3 years ago
3 0

72 1/10% = 72.1% then you can put it in decimal form by moving two decimal places left .721 and take away the percentage symbol, the word "of" in this case means you're multiplying so it will be .721 x 19 and that's how you get your answer

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Find the derivative of the following. please show the steps when you answer :1) f(x) = 8xe^x2) y= 5xe^x^43) f(x)= x^8+5/x4) f(t)
borishaifa [10]

Answer:

Since,

\frac{d}{dx}x^n = nx^{n-1}

\frac{d}{dx}(f(x).g(x)) = f(x).\frac{d}{dx}(g(x)) + g(x).\frac{d}{dx}(f(x))

\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x).f'(x) - f(x) g'(x)}{(g(x))^2}

1) y = 8x e^x

Differentiating with respect to x,

\frac{dy}{dx}=8( x \times e^x + e^x) = 8(xe^x + e^x) = 8e^x(x+1)

2) y = 5x e^{x^4}

Differentiating w. r. t x,

\frac{dy}{dx}=5(x\times 4x^3 e^{x^4}+e^{x^4})=5e^{x^4}(4x^4+1)

3) y = x^8 + \frac{5}{x^4}

Differentiating w. r. t. x,

\frac{dy}{dx}=8x^7 - \frac{5}{x^5}\times 4 = 8x^7 - \frac{20}{x^5}=\frac{8x^{12}-20}{x^5}

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Differentiating w. r. t. t,

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Differentiating w. r. t. p,

g'(p) = p\frac{1}{2p+3}(2) + \ln(2p+3) = \frac{2p}{2p+3}+\ln(2p+3)

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Differentiating w. r. t. t,

\frac{dz}{dt}=7(te^{6t}+e^{5t})^6 ( 6te^{6t}+e^{6t} + 5e^{5t})

7) w =\frac{2y + y^2}{7+y}

Differentiating w. r. t. y,

\frac{dw}{dy} = \frac{(7+y)(2+2y)-(2y+y^2)}{(7+y)^2} = \frac{14 + 2y + 14y +2y^2 - 2y - y^2}{(7+y)^2}=\frac{14+14y+y^2}{(7+y)^2}

7 0
3 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
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mamaluj [8]

Answer:

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This is a simple interest problem.

The simple interest formula is given by:

E = P*I*t

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After t years, the total amount of money is:

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We have to find P for which T = 3000. So

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P = 2790.7

He should pay $2,790.7.

8 0
3 years ago
How do you find absolute extrema for a function?<br> f(x)= (8+x)/(8-x); Interval of [4,6]
Mkey [24]
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so, from zeroing out the derivative we get no critical points there, from the denominator we get x = 8, but can't use it because f(x) is undefined.

therefore, we settle for the endpoints, 4 and 6,

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doing a first-derivative test, we see the slope just goes up at both points and in between, but the highest is f(6), so the absolute maximum is there, while we can take say f(4) as the only minimum and therefore the absolute minumum as well.
3 0
3 years ago
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