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musickatia [10]
3 years ago
10

1.The equation of line CD is (y−3) = − 2 (x − 4). What is the slope of a line perpendicular to line CD? 2.The equation of line A

B is y = 2x + 4. Write an equation of a line parallel to line AB in slope-intercept form that contains point (3, −2). 3.The equation of line QR is y = negative 1/2x + 1. Write an equation of a line perpendicular to line QR in slope-intercept form that contains point (5, 6). 4.Line WX contains (−1, 2) and (4, 12) Line YZ contains points (5, 8) and (−2, −6). Lines WX and YZ are 5.Line CD contains points A (4, −7) and B (4, 8). The slope of line CD
Mathematics
2 answers:
Helga [31]3 years ago
8 0

Answer:

m= y2-y1 ÷ X2 - X1

Step-by-step explanation:

gogolik [260]3 years ago
5 0
So here's how to do it,
slope intercept form is y=mx+b
m is the slope, b is the y- intercept, X is your unknown variable.

you need to solve to get y by its self to get your slope.

for perpendicular lines flip the slope

ex 1, smy=2x+6 --- slope of a perpendicular line will be 1/2

ex 2, y=3/2+8 ---8 perpendicular lien would have a slope of 2/3

parallel lines have the same slope so any equation with with the same slope will be parallel if it has the same degree.

ex. y=2x+14 is parallel to y= 2x+56

slope can be found using the following formula if you have two sets of points on the same line

m= y2-y1 ÷ X2 - X1
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GrogVix [38]

Answer:

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Step-by-step explanation:

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3 0
3 years ago
Given that tan theta = -4/7, and 270° < theta < 360°, what is the exact value of sec theta​
aleksley [76]

Answer:

sec Θ = \frac{\sqrt{65} }{7}

Step-by-step explanation:

Using the trigonometric identity

sec²Θ = tan²Θ + 1

Since 270° < Θ < 360° ← that is fourth quadrant, then

sec Θ > 0, thus

sec²Θ = (- \frac{4}{7})² + 1 = \frac{16}{49} + 1 = \frac{65}{49}, then

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6 0
3 years ago
What is the answer to the question 25% of what is 21?
g100num [7]
If we call the number "x" then:
x*25%=21
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8 0
3 years ago
Read 2 more answers
If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?
kondaur [170]

Equation of the circle is (x+6)^{2}+(y+10)^{2}=20.

Solution:

The endpoints of the diameter of a circle are (–8, –6) and (–4, –14).

Center of the circle = Mid point of the diameter

Mid point formula:

$P(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)

Here, x_1=-8, y_1=-6, x_2=-4, y_2=-14

$P(x, y) =\left(\frac{-8-4}{2}, \frac{-6-14}{2}\right)

$P(x, y) =\left(\frac{-12}{2}, \frac{-20}{2}\right)

$P(x, y) =(-6, -10)

Center of the circle = (–6, –10)

Radius is the distance between center and any endpoint of the diameter.

To calculate the radius using distance formula.

r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Here, x_1=-6, y_1=-10, x_2=-8, y_2=-6

r=\sqrt{\left(-8-(-6)\right)^{2}+\left(-6-(-10)}\right)^{2}}

r=\sqrt{(-8+6)^{2}+(-6+10)^{2}}

r=\sqrt{(-2)^{2}+(4)^{2}}

r=\sqrt{20} units

The standard form of the equation of a circle is

(x-a)^{2}+(y-b)^{2}=r^{2}, where (a, b) are center and r is the radius.

Here, center = (–6, –10) and r=\sqrt{20}

(x-(-6))^{2}+(y-(-10))^{2}={(\sqrt{20})} ^{2}

(x+6)^{2}+(y+10)^{2}=20

Equation of the circle is (x+6)^{2}+(y+10)^{2}=20.

4 0
3 years ago
Pls, help me with this function table in the screenshot below...
Y_Kistochka [10]

Answer:

-17

Step-by-step explanation:

Every number x goes up, the f(x) goes up by 9. -8+9=1 and 1+9=10. Therefore if the number goes down, you subtract. -8-9=-17

7 0
3 years ago
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