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diamong [38]
3 years ago
5

Will mark brainliest if correct :)

Mathematics
1 answer:
Finger [1]3 years ago
8 0

Answer:

d = 12

c = 4

so I think d > c

Step-by-step explanation:

1 x 12 = 12

0.75 x 4 = 3

12 + 3 = 15

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Solve for f.<br><br> 8 = 2f + 2
tiny-mole [99]

Answer:

f = 3

Step-by-step explanation:

8 = 2f + 2

-2         -2

6 = 2f

/2    /2

3 = f

5 0
3 years ago
In the diagram below, f(x)=x^3+2x^2 is graphed. Also graphed is g(x), the result of a translation of f(x). Determine an equation
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Answer:

g(x)=x^3+2x^2 - 4

Step-by-step explanation:

See attached worksheet

6 0
2 years ago
Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is con
monitta

Let f(x)=e^{-1/x}. Then f'(x)=\frac1{x^2}e^{-1/x}>0 for all x\ge2, so f is strictly increasing. As x\to\infty, e^{-1/x}\to e^0=1, so f is bounded above by 1. This is to say,

e^{-1/x}

and the integral of \frac1{x^2} converges over the same domain, so this integral must also converge by comparison.

We have, by setting y=-\frac1x,

\displaystyle\int_2^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx=\int_{-1/2}^0e^y\,\mathrm dy=e^0-e^{-1/2}=1-\frac1{\sqrt e}

8 0
3 years ago
1/12 the fraction divided by 30
nexus9112 [7]
1/12 ÷30 is equal to 1/360 or in decimal form is a repetitive decimal which is .2777777777777777777777....
7 0
3 years ago
Read 2 more answers
Find the derivative.
Aleksandr [31]

Answer:

Using either method, we obtain:  t^\frac{3}{8}

Step-by-step explanation:

a) By evaluating the integral:

 \frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du

The integral itself can be evaluated by writing the root and exponent of the variable u as:   \sqrt[8]{u^3} =u^{\frac{3}{8}

Then, an antiderivative of this is: \frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}

which evaluated between the limits of integration gives:

\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}

and now the derivative of this expression with respect to "t" is:

\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

g(x)=\int\limits^x_a {f(t)} \, dt

is continuous on [a,b], differentiable on (a,b) and  g'(x)=f(x)

Since this this function u^{\frac{3}{8} is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}

5 0
3 years ago
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