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3 years ago

What is the domain of the function g(x)=5^2x

Mathematics

1 answer:

IgorC [24]3 years ago

4 0

$g(x)=5^{2x}$
The domain:
All real numbers

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What is the domain of the function 1x)= 2

Gnesinka [82]

Answer:

According to me, the answer is

Step-by-step explanation:

all real numbers

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3 years ago

1. In the diagram, AB is parallel to CD.

stira [4]

Answer:

x= 72°

reason: this is because it is a corresponding angle. Corresponding angles are equal to another

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3 years ago

What is the distance between (2,-6) and (-4,8)

tangare [24]

The formula for distance is:

$d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}$

Use this formula to solve your problem where:

1 = (2, -6)

2 = (-4, 8)

$d = \sqrt{(-4 - 2)^2 + (8 + 6)^2}\\\\d = \sqrt{(-6)^2 + (14)^2}\\\\d = \sqrt{36 + 196}\\\\d = \sqrt{232} = 15.232$

The distance between (2, -6) and (-4, 8) is 15.232 units.

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3 years ago

Stan took 25 classes during college. The Venn diagram below shows classes Stan studied for, classes Stan

klio [65]

Answer:

Studied for the class:

Passed: 12

Didn't: 0

Didn't study for the class:

Passed: 2

Didn't: 11

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3 years ago

For the function below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subin

Vlad [161]

Answer:

See below

Step-by-step explanation:

We start by dividing the interval [0,4] into n sub-intervals of length 4/n

$[0,\displaystyle\frac{4}{n}],[\displaystyle\frac{4}{n},\displaystyle\frac{2*4}{n}],[\displaystyle\frac{2*4}{n},\displaystyle\frac{3*4}{n}],...,[\displaystyle\frac{(n-1)*4}{n},4]$

Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.

Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)

$\displaystyle\frac{4}{n}f(\displaystyle\frac{1*4}{n})+\displaystyle\frac{4}{n}f(\displaystyle\frac{2*4}{n})+...+\displaystyle\frac{4}{n}f(\displaystyle\frac{n*4}{n})=\\\\=\displaystyle\frac{4}{n}((\displaystyle\frac{1*4}{n})^2+3+(\displaystyle\frac{2*4}{n})^2+3+...+(\displaystyle\frac{n*4}{n})^2+3)=\\\\\displaystyle\frac{4}{n}((1^2+2^2+...+n^2)\displaystyle\frac{4^2}{n^2}+3n)=\\\\\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12$

but

$1^2+2^2+...+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}$

so the upper sum equals

$\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12=\displaystyle\frac{4^3}{n^3}\displaystyle\frac{n(n+1)(2n+1)}{6}+12=\\\\\displaystyle\frac{4^3}{6}(2+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2})+12$

When $n\rightarrow \infty$ both $\displaystyle\frac{3}{n}$ and $\displaystyle\frac{1}{n^2}$ tend to zero and the upper sum tends to

$\displaystyle\frac{4^3}{3}+12=\displaystyle\frac{100}{3}$

8 0

4 years ago