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3 years ago

May someone help, please ? Thank you ♥️

Mathematics

2 answers:

xeze [42]3 years ago

4 0

The answer is c. A right triangle is one where a^2 + b^2 = c^2, where c is the longest side.

Brums [2.3K]3 years ago

3 0

C. is the correct answer.

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14. Find the solution of the initial value problem y′′ + y = F(t), y(0) = 0, y′(0) = 0, where F(t) = ⎧⎪⎨⎪⎩ At, 0≤ t ≤ ????, A(2?

Vinvika [58]

Yxgxgggffggyygfccccgyu

3 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

What is -8 square root of 9 when simplified

gregori [183]

Answer:

$\large\boxed{-8\sqrt9=-24}$

Step-by-step explanation:

$\sqrt{a}=b\iff b^2=a\\\\\sqrt9=3\ \text{because}\ 3^2=9\\\\-8\sqrt9=-8(3)=-24$

7 0

3 years ago

Find the equation of a line that contains the points (3,7) and (-6, 4). Write the equation in slope-intercept form, using

NikAS [45]

Answer:

$y=\frac{1}{3} x+6$

Step-by-step explanation:

$(3,7)(-6,4)$

Step 1. Find the slope (by using the slope-formula)

m = slope

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{4-7}{-6-3}$

$m=\frac{-3}{-9}$

$m=\frac{3}{9}$

$m=\frac{1}{3}$

Step 2. Write the equation (using the slope and the points)

Here's how to do it:

Slope-intercept Formula $y=mx+b$ whrere m = slope and b = y-intercept

Plug in the slope into the Slope-intercept Formula

$y=\frac{1}{3} x+b$

Find the y-intercept (b) by using a point and substituting their x and y values

$y=\frac{1}{3} x+b$

Point: (3, 7)

$7=\frac{1}{3} (3)+b$

$7=1+b$

$b=7-1$

$b=6$

Step 3. Write the equation in Slope-intercept form

$y=mx+b$

$y=\frac{1}{3} x+6$

3 0

3 years ago

What’s this guys ? I’ve been stuck on this ‍♀️

Harrizon [31]

SOLUTION:

Let's establish the formula for a cylinder as displayed below:

Let volume of cylinder = V

V = ( Pi )r^2h

Now let's substitute the values from the problem into the formula to find the volume.

V = ?

r = 8

h = 4

V = ( Pi )( 8 )^2( 4 )

V = ( Pi )( 64 )( 4 )

V = ( Pi )( 256 )

V = 256( Pi )

FINAL ANSWER:

Therefore, the answer is:

C. 256( Pi ) units^3

Hope this helps! :)
Have a lovely day! <3

3 0

3 years ago