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4 years ago

Please answer this question fast in two muintues

Mathematics

2 answers:

Minchanka [31]4 years ago

5 0

Answer:

Step-by-step explanation:

The vertex is where the 2 rays meet, or the corner of the angle

The vertex is W

djyliett [7]4 years ago

4 0

Answer:

Step-by-step explanation:

W is the vertex, you can see the letter above the angle

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Which of the following are exterior angles?Check all that apply. Help asap plz

ddd [48]

Angle 6, angle 4, angle 3, and angle 2 are all exterior angles.

Hope this helps!

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4 years ago

Read 2 more answers

True or False: If a chi-square test is significant (i.e., there is an association between the variables), we should look at the

AnnyKZ [126]

Answer: True.

Step-by-step explanation:

A conditional distribution in this case is a probability distribution for a sub-population. In other words, it represent the probability that a randomly selected item in a sub-population has a characteristic we’are interested in .

Partitioning the popular action assist us to see if the sub-population contains the characteristics we ere interested in I.e . Association

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3 years ago

Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

$u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}$

$u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}$

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

$u'x+u=\frac{u+u^2}{3+u^2}$

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

$u'x=\frac{u+u^2-3u-u^3}{3+u^2}$

$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

$\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}$

Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0

3 years ago

If a line segment is perpendicular to another line segment, it is the bisector of that line segment.

sp2606 [1]

False. Perpendicular lines run beside each other not through each other.

7 0

4 years ago

Indicate in standard form the equation of the line passing through the given points, writing the answer in the equation

Kryger [21]

Answer:

$y + x - 6 = 0$

Step-by-step explanation:

Given

$M(0,6) \ and \ N(6,0)$

Required

Find the equation of the line

First, the slope of the line has to be calculated using the following formula;

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

$where\ (x_1,y_1) = (0,6) \ and \ (x_2,y_2) = (6,0)$

So, the equation becomes

$m = \frac{0 - 6}{6 - 0}$

$m = \frac{-6}{6}$

$m = -1$

The equation of the line can then be calculated using

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

$Using \ m = \frac{y - y_1}{x - x_1}$

$-1 = \frac{y - 6}{x -0}$

$-1 = \frac{y - 6}{x}$

Multiply both sides by x

$-1 * x = \frac{y - 6}{x} * x$

$-x = y - 6$

Add x to both sides

$x -x = y - 6 + x$

$0 = y - 6 + x$

Reorder

$y + x- 6 = 0$

$Using \ m = \frac{y - y_2}{x - x_2}$

$-1 = \frac{y - 0}{x - 6}$

$-1 = \frac{y}{x - 6}$

Multiply both sides by x - 6

$-1 * (x-6) = \frac{y}{x - 6} * (x-6)$

$-1 * (x-6) = y$

$-x+6 = y$

Add x - 6 to both sided

$x - 6 -x+6 = y +x - 6$

$0 = y + x - 6$

$y + x - 6 = 0$

Hence, the equation of the line is $y + x - 6 = 0$

3 0

3 years ago