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4 years ago

Differentiate 2sin(5x-5)

Mathematics

1 answer:

Ainat [17]4 years ago

4 0

$f(x)=2\sin(5x-5)\\
f'(x)=2\cos(5x-5)\cdot5=10\cos(5x-5)$

Read more about proportions at:

brainly.com/question/843074

5 0

2 years ago

Using the point-slope formula, what is the equation of the line that passes through the points (1,5) and (0, 0)?

DerKrebs [107]

Answer:

y =5x

Step-by-step explanation:

$y_{1}-y_{2} = m(x_{1} - x_{2} )$

(5 - 0) = m(1 - 0)

5 = m(1)

5 = m

$y_{1}-y_{2} = m(x_{1} - x_{2} )$

y - 5 = 5(x - 1)

y - 5 = 5x - 5

y = 5x -5 + 5

y = 5x

7 0

3 years ago

The following data shows the number of home runs hit by the top 12 home run hitters in Major League Baseball during the 2011 sea

KATRIN_1 [288]

Answer:

d. 25.25.

Step-by-step explanation:

A whisker plot is a type of box plot which is graphical representation of five number summary. It is used for explanatory data analysis. The baseball league has data set whose median is 45. When the outliner are present in data set the median measures central tendency.

8 0

4 years ago

Please help i’m so stuck i can’t do calc for the life of me

KATRIN_1 [288]

Answer:

$-3x^2+29x-150+\frac{946x^2-341x-756}{x^3+6x^2-3x-5}$ is the answer. See steps below.

Step-by-step explanation:

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}\\\\-3x^5+11x^4+33x^3-26x^2-36x-6\\\\\mathrm{and\:the\:divisor\:}x^3+6x^2-3x-5:\\\\\mathrm{Quotient}=\frac{-3x^5}{x^3}=-3x^2\\\\\mathrm{Multiply\:}x^3+6x^2-3x-5\mathrm{\:by\:}-3x^2\:\:\rightarrow\:\:-3x^5-18x^4+9x^3+15x^2\\\\\mathrm{Subtract\:}-3x^5-18x^4+9x^3+15x^2\mathrm{\:from\:}-3x^5+11x^4+33x^3-26x^2-36x-6\mathrm{\:to\:get\:new\:remainder}.\\\\\mathrm{Remainder}=29x^4+24x^3-41x^2-36x-6$

$=-3x^2+\frac{29x^4+24x^3-41x^2-36x-6}{x^3+6x^2-3x-5}$

Repeat the steps and you will reach a point where no further division is possible.

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5 0

3 years ago

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

3 0

3 years ago