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3 years ago

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What is the equation in point-slope form of the line passing through (4, 0) and (2, 6)? (5 points) y = 4x − 2 y = 2x − 4 y = −3(

x − 4) y = 3(x + 4)

2 answers:

tester [92]3 years ago

8 0

Answer:

Third option: $y=-3(x-4)$

Step-by-step explanation:

The equation of the line in Point-Slope form is:

$y-y_1=m(x-x_1)$

Where "m" is the slope and $(x_1,y_1)$ is a point on the line.

Given the points (4, 0) and (2, 6), we can find the slope with this formula:

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

Substituting values, we get:

$m=\frac{0-6}{4-2}=-3$

Finally, substituting the slope and the point (4,0) into $y-y_1=m(x-x_1)$ , we get:

$y-0=-3(x-4)$

$y=-3(x-4)$

DIA [1.3K]3 years ago

6 0

For this case we have that by definition, the point-slope equation of a line is given by:

$y-y_ {0} = m (x-x_ {0})$

We have the following points:

$(x_ {1}, y_ {1}) :( 2,6)\\(x_ {2}, y_ {2}) :( 4,0)\\m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {0-6} {4-2} = \frac {-6} {2} = -3$

We chose a point:

$(x_ {0}, y_ {0}) :( 4,0)$

Substituting in the equation we have:

$y-0 = -3 (x-4)\\y = -3 (x-4)$

Finally, the equation is: $y = -3 (x-4)$

Answer:

OPTION C

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Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

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Annette [7]

Answer:

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Step-by-step explanation:

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Genrish500 [490]

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We rewrite the expression using the lowest common index of 6, then:

$5 ^ {\frac {1} {3}} * 2 ^ {\frac {1} {2}} =$

We rewrite the terms in an equivalent way:

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We rewrite the expression using the property mentioned:

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We combine using the product rule for radicals:

$\sqrt [n] {a} * \sqrt [n] {b} = \sqrt [n] {ab}$

So:

$\sqrt [6] {5 ^ 2 * 2 ^ 3} =\\\sqrt [6] {25 * 8} =\\\sqrt[6]{200}$

ANswer:

Option b

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