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3 years ago

3 3/5 + 6 3/10 please

1 answer:

3 0

In order to solve first convert both values into improper fractions:
18/5+63/10
second, convert both values to fractions with a base of ten:
36/10+63/10
finally, add the numerators:
99/10 is the final answer

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What is 0+0 plz help me o am in college and need help

s2008m [1.1K]

The answer is 0

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7 0

3 years ago

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Nichole bought a gift for her friend. She

OlgaM077 [116]

Answer:

10 by 5 =50

10 by 5=50

10 by 2=20

2 by 5=10

100+40+20=160

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3 years ago

Divide using long division. <br><br> <img src="https://tex.z-dn.net/?f=%284x%5E%7B2%7D%20-5x%2B3%29" id="TexFormula1" title="(4x

frez [133]

Answer:

The quotient = 4x - 17 and the remainder = 54

Step-by-step explanation:

$\frac{4x^{2}-5x+3 }{x+3}=4x+\frac{-17x+3}{x+3}$

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6 0

2 years ago

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Over a 3.5 year period, 27.5 inches of snow fell in Jackson City. What was the average yearly snowfall in Jackson Cit

Ilya [14]

27.5 / 3.5 = 7.86; The average yearly snowfall was 7.86 inches.

7 0

2 years ago

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