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3 years ago

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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Subtracting a negative is the same as....

erik [133]

Same as adding together

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2 years ago

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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,

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

4) $f(t) = te^{11}-6t^5$

Differentiating w. r. t. t,

$f'(t) = e^{11} - 30t^4$

5) $g(p) = p\ln(2p+3)$

Differentiating w. r. t. p,

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

6) $z = (te^{6t}+e^{5t})^7$

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

30 POINTS Type the correct answer in the box. Use numerals instead of words.

olga55 [171]

Answer:

15 rubber bands per day

Step-by-step explanation:

5 0

3 years ago

3. Bob the Builder wants to earn an annual rate of 10% on his investments,

mamaluj [8]

Answer:

He should pay $2,790.7.

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

$E = P*I*t$

In which E is the amount of interest earned, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time, in years.

After t years, the total amount of money is:

$T = E + P$

In this question:

Rate of 10%, so I = 0.1.

9 months, so $t = \frac{9}{12} = 0.75$

How much should he pay for a note that will be worth $3,000 in 9 months?

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

$T = E + P$

$3000 = E + P$

$E = 3000 - P$

Then

$E = P*I*t$

$3000 - P = P*0.1*0.75$

$1.075P = 3000$

$P = \frac{3000}{1.075}$

$P = 2790.7$

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]

$\bf f(x)=\cfrac{8+x}{8-x}\implies \cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{1(8-x)-(8+x)(-1)}{(8-x)^2}}\implies \cfrac{dy}{dx}=\cfrac{16}{(8-x)^2}$

now, we get critical points from zeroing out the derivative, and also from zeroing out the denominator, but those at the denominator are critical points where the function is not differentiable, namely a sharp spike or cusp or an asymptote.

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,

f(4) =3 and f(6) = 7

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