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3 years ago

I need help w this question

Mathematics

1 answer:

ollegr [7]3 years ago

3 0

Answer:

360 degrees

Step-by-step explanation:

Remember that triangles have a sum of 180 degrees, and that quadrilaterals have 360 degrees.

You can either use that part, or actually solve for it.

The equation for the sum of the interior angles is:

$sum=180(n-2)$ [which "n" is the number of sides]

So, if we add in the number of sides to the equation, it changes to:

$180(4-2)=180(2)$ , which equals to 360 degrees.

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mash [69]

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3 years ago

Please help, this will just take a minute of your time ❤︎ (literally)<br>(Discriminant Worksheet)

igomit [66]

Answer:

$\large\boxed{x=-\dfrac{7-\sqrt{97}}{4}\ \vee\ x=-\dfrac{7+\sqrt{97}}{4}}$

Step-by-step explanation:

$ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\\text{if}\ \Delta0,\ \text{then an equation has two solutions:}\ x=\dfrac{-b\pm\sqrt\Delta}{2a}\\\\\text{We have:}\\\\-2x^2-7x+10=4\qquad\text{subtract 4 from both sides}\\\\-2x^2-7x+6=0$

$a=-2,\ b=-7,\ c=6\\\\\Delta=(-7)^2-4(-2)(6)=49+48=97>0\\\\\sqrt\Delta=\sqrt{97}\\\\x=\dfrac{-(-7)\pm\sqrt{97}}{2(-2)}=\dfrac{7\pm\sqrt{97}}{-4}=-\dfrac{7\pm\sqrt{97}}{4}$

6 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:

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