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3 years ago

Which number line correctly shows 0.8 + 0.3?

Mathematics

2 answers:

GenaCL600 [577]3 years ago

8 0

Answer:

B. Is the correct way to show the problem.

Step-by-step explanation:

A: The number line is useless. It shows working backwards from the answer. You want to show how the answer was obtained. + goes right - goes left. Not A. This actually shows 1.1 - 0.8 - 0.3

C: C shows 0.8 - 0.3. Remember + goes right - goes left.

D: D is wrong. It shows 0.3 to 1.1 and then 0.3 back to 0.8. You are not doing that.

worty [1.4K]3 years ago

6 0

Answer:

Step-by-step explanation:

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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:

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

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