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kati45 [8]
3 years ago
13

Type the correct answer in the box. Use numerals instead of words. What is the solution for x in the equation? 10x − 4.5 + 3x =

12x − 1.1 x =
Mathematics
1 answer:
kakasveta [241]3 years ago
5 0
10x - 4.5 + 3x = 12x - 1.1
13x - 4.5 = 12x - 1.1
13x - 12x - 4.5 = 12x - 12x - 1.1
1x - 4.5 = -1.1
1x - 4.5 + 4.5 = -1.1 + 4.5
1x = 3.4
X = 3.4
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The answer is in the image! The interval notation is [-5,4)

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Notice that on the linee graph it shows a open circle on 4 and a closed dot on -5... This is because the number can be greater than -5 or equal to(this is represented by the brackets) and on 4 it can only be less than 4 and not equal to it (Represented by parenthesis)...

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Read 2 more answers
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
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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}

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3 years ago
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