20 points!!!Please help HURRY!!!

Mathematics

1 answer:

navik [9.2K]2 years ago

5 0

Answer:

Definitely not!

Step-by-step explanation:

It was reflected across the x axis, resized and moved in x and y directions, but since the shape ABCD is symmetrical across the y axis, a reflection across y axis won’t change anything!

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A Write an equivalent equation that does NOT contain decimals.

vampirchik [111]

0.5x - 0.1 = -2.9

When you multiply each side of an equation by the same number,
you don't change the equation or its solution. If you want to get rid
of the decimals, you could multiply each side by 10. Then the
equation is

 5x - 1 = -29

You should be able to solve that without help.
But I'm about to take your points, so here's
how to solve that equation:

 5x - 1 = -29

Add 1 to each side: 5x = -28

Divide each side by 5 : x = -28 / 5

Simplify the solution: x = -5 and 4/5 .

Again ... this is the same solution as the original equation
if we hadn't gotten rid of the decimals.

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3 years ago

What is the standard deviation of 98.21

antiseptic1488 [7]

Answer: 0.61

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find integra of xlnxdx

Mademuasel [1]

Answer:

$\dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}$

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.

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

Increase the power by 1, then divide by the new power.

Given indefinite integral:

$\displaystyle \int x \ln x \:\: \text{d}x$

To integrate the given integral, use Integration by Parts:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

$\text{Let }u=\ln x \implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{x}$

$\text{Let }\dfrac{\text{d}v}{\text{d}x}=x \implies v=\dfrac{1}{2}x^2$

Therefore:

$\begin{aligned}\displaystyle \int u \dfrac{dv}{dx}\:dx & =uv-\int v\: \dfrac{du}{dx}\:dx\\\\\implies \displaystyle \int x \ln x\:\:\text{d}x & = \ln x \cdot \dfrac{1}{2}x^2-\int \dfrac{1}{2}x^2 \cdot \dfrac{1}{x}\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x -\int \dfrac{1}{2}x\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}\end{aligned}$