Round each number to the nearest tenth.

Mathematics

1 answer:

andreev551 [17]3 years ago

7 0

Answer:For the first one 26.73 would turn into 27 and 89.42 would stay 89

Step-by-step explanation: Any number that is 5 or greater can be used to round any number.

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A systematic relationship between two variables in which a change in one implies a corresponding change in the other is called _

Sedaia [141]

A systematic relationship between two variables in which a change in one implies a corresponding change in the other is called proportional relationship.

Two variables are said to be proportional if an increase in one variable leads to an increase in the other variable and vice versa.

Let us assume that y and x are proportional to each other. Hence:

y ∝ x

y = kx

Where k is the constant of proportionality.

Find out more about proportional relationship at: brainly.com/question/25718473

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2 years ago

Please help 15 POINTS PROBLEM BELOW

Anit [1.1K]

be sure to use PEMDAS so first use the distributive property then work your way from there.

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3 years ago

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Evaluate the following integral (Calculus 2) Please provide step by step explanation!

Step2247 [10]

Answer:

$\displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x=-\dfrac{2}{x+1}+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given integral:

$\displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x$

Factor the denominator:

$\begin{aligned}\implies x^2+2x+1 & = x^2+x+x+1\\& = x(x+1)+1(x+1)\\& = (x+1)(x+1)\\& = (x+1)^2\end{aligned}$

$\implies \displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x=\int \dfrac{2}{(x+1)^2}\:\:\text{d}x$

$\textsf{Apply exponent rule} \quad \dfrac{1}{a^n}=a^{-n}$

$\implies \displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x=\int 2(x+1)^{-2}\:\:\text{d}x$

$\boxed{\begin{minipage}{4 cm}\underline{Integrating $ax^n$}\\\\$\displaystyle \int ax^n\:\text{d}x=\dfrac{ax^{n+1}}{n+1}+\text{C}$\end{minipage}}$

Use Integration by Substitution:

$\textsf{Let }u=(x+1) \implies \dfrac{\text{d}u}{\text{d}x}=1 \implies \text{d}x=\text{d}u}$

Therefore:

$\begin{aligned}\displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x & = \int 2(x+1)^{-2}\:\:\text{d}x\\\\& = \int 2u^{-2}\:\:\text{d}u\\\\& = \dfrac{2}{-1}u^{-2+1}+\text{C}\\\\& = -2u^{-1}+\text{C}\\\\& = -\dfrac{2}{u}+\text{C}\\\\& = -\dfrac{2}{x+1}+\text{C}\end{aligned}$