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2 years ago

Find the missing side length, indicate whether the side lengths form a Pythagorean triple, and identify the correct explanation.

Mathematics

2 answers:

saw5 [17]2 years ago

4 0

Step-by-step explanation:

Pythagorean Formula = a2 + b2 = C

which is a = Side of triangle

b = Side of Triangle

c = Hypotenuse

Bess [88]2 years ago

3 0

Answer:

Step-by-step explanation:

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A bag contains 6 red balls and 3 blue balls. 4 balls are selected at random. find the probability of selecting 4 red balls.

geniusboy [140]

Assuming that the balls are not replaced;

$\frac{6}{9}* \frac{5}{8}* \frac{4}{7}* \frac{3}{6}$ = 5/42

(the numbers are decreasing in the denominator as you are presumably taking out the red balls. In the denominator, there is one less ball each time since you took out a red one).

Therefore, the probability of selected 4 red balls consecutively without replacement is 5/42.

Hope I Helped :)

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3 years ago

A cube has a surface area of 486 square meters. What is the volume of the cube.

Oksana_A [137]

V= s^3
a=486/6=81
s=sqrt(81)=9
v=9^3=729
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2 years ago

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Follow the process of completing the square to solve 2x2 + 8x - 12 = 0. After adding B2 to both sides of the equation in step 4,

Katena32 [7]

2x^2 + 8x - 12 = 0..divide by 2
x^2 + 4x - 6 = 0
x^2 + 4x = 6...add 4 to both sides of the equation
x^2 + 4x + 4 = 6 + 4
(x + 2)^2 = 10....<== ur constant is 10
x + 2 = (+-)sqrt 10
x = -2 (+ - ) sqrt 10

x = -2 + sqrt 10
x = -2 - sqrt 10

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2 years ago

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G evaluate the given integral by changing to polar coordinates. e−x2 − y2 da d where d is the region that lies to the left of th

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2 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

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2 years ago