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4 years ago

X intercepts for the equation y=-3(x-2)^2+1

Mathematics

1 answer:

nignag [31]4 years ago

5 0

Answer:

Step-by-step explanation:

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What is the discount on a stethoscope that is marked down from $300 to $250?

labwork [276]

To find this answer you simply divide 250$ by 300$ you would then get 83% that is not the answer though you then subtract 83 from 100 and get 17 and that is the discount

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3 years ago

Which table represents a linear function?

igomit [66]

Answer:

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Step-by-step explanation:

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3 years ago

carmen participated in a read-a-thon.MR.cole pledge $4.00 per book and gave carmen $44.00.How many books did carmen read

finlep [7]

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Step-by-step explanation:

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4 years ago

(12z+7z-15) divide (3z^3-2z-8)

yarga [219]

(12z + 7z -15) / (3z^3 - 2z - 8)

Solution: 19z − 15 / 3z^3 − 2z −8

Steps:

(12z + 7z -15) / (3z^3 - 2z - 8)

12z + 7z - 15 = 19z - 15

12z + 7z - 15

Add Similar Elements:

12z + 7z = 19z

19z - 15

Hence, Answer, (12z + 7z -15) / (3z^3 - 2z - 8) ===================> 19z − 15 / 3z^3 − 2z −8

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4 0

4 years ago

Find the integrate of 2x/sqrt(1-x^2) dx from 0 to 1/2

Ludmilka [50]

Compute the definite integral:

$\large\begin{array}{l}\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}\,dx}\\\\ =\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{-1}{\sqrt{1-x^2}}\cdot (-2x)\,dx\qquad\quad(i)} \end{array}$

$\large\begin{array}{l} \textsf{Substitute}\\\\ \mathsf{1-x^2=u~~\Rightarrow~~-2x\,dx=du}\\\\\\ \textsf{Finding the new limits of integration:}\\\\ \begin{array}{lcl} \textsf{When }\mathsf{x=0}&~\Rightarrow~&\mathsf{u=1-0^2}\\\\ &&\mathsf{u=1}\\\\\\ \textsf{When }\mathsf{x=\dfrac{1}{2}}&~\Rightarrow~&\mathsf{u=1-\left(\dfrac{1}{2}\right)^2}\\\\ &&\mathsf{u=1-\dfrac{1}{4}}\\\\ &&\mathsf{u=\dfrac{3}{4}} \end{array} \end{array}$

$\large\begin{array}{l} \textsf{Then (i) becomes}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}}\frac{-1}{\sqrt{u}}\,du}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}}\frac{-1}{u^{\frac{1}{2}}}\,du}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}} (-1)\cdot u^{-\frac{1}{2}}\,du}\\\\ =\mathsf{(-1)\cdot \dfrac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg|_{1}^{\frac{3}{4}}} \end{array}$

$\large\begin{array}{l} =\mathsf{(-1)\cdot \dfrac{~~u^{\frac{1}{2}}~~}{\frac{1}{2}}\bigg|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{(-1)\cdot 2u^{\frac{1}{2}}\Big|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{-2\cdot \sqrt{u}\Big|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{-2\cdot \left(\sqrt{\dfrac{3}{4}}-\sqrt{1}\right)} \end{array}$

$\large\begin{array}{l} =\mathsf{-2\cdot \left(\dfrac{\sqrt{3}}{2}-1\right)}\\\\ =\mathsf{-\diagup\!\!\!\! 2\cdot \dfrac{\sqrt{3}}{\diagup\!\!\!\! 2}+(-2)\cdot (-1)}\\\\ =\mathsf{-\sqrt{3}+2}\\\\ =\mathsf{2-\sqrt{3}} \end{array}$

$\large\begin{array}{l} \boxed{\begin{array}{c}\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}\,dx=2-\sqrt{3}} \end{array}}\qquad\checkmark \end{array}$

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$\large\textsf{I hope it helps.}$

Tags: definite integral integrate limits function irrational square root sqrt fraction composite substitution integral calculus

4 0

4 years ago