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Zinaida [17]
3 years ago
14

Solve the equation. 2/3 -4c+7/2 = -9x+5/6

Mathematics
1 answer:
Airida [17]3 years ago
8 0

Answer:

The value of X is -2/3

Step-by-step explanation:

edge2021

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Step-by-step explanation:

y = 4x - 1 \\ m = 4 \\ c =  - 1

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In a certain neighborhood, there's a sagging power line between two utility poles. The utility poles are 50 feet tall and 120 fe
e-lub [12.9K]
<h3>Answer:  33.75 feet</h3>

In fraction form, this value is equal to 135/4

33.75 ft is equivalent to 33 ft, 9 inches.

===============================================

Explanation:

Refer to the diagram below.

The key point to start with is point H, which is the vertex of the parabola.

Recall that vertex form is

y = a(x-h)^2 + k

What we'll do is plug in the vertex (h,k) = (60,30) which is the location of point H. We'll also plug in (0,45) which is the y intercept, aka the location of point C.

So,

y = a(x-h)^2 + k

y = a(x-60)^2 + 30 .... plug in vertex

45 = a(0-60)^2 + 30 .... plug in y intercept coordinates

45 = a(-60)^2 + 30

45 = a(3600) + 30

45 = 3600a + 30

45-30 = 3600a

3600a = 15

a = 15/3600

a = 1/240

This then means:

y = a(x-h)^2 + k

y = (1/240)(x-60)^2 + 30

This is the equation of our parabola. Plug in x = 30 to determine the height of point K

y = (1/240)(x-60)^2 + 30

y = (1/240)(30-60)^2 + 30

y = (1/240)(-30)^2 + 30

y = (1/240)(900) + 30

y = 15/4 + 30

y = 15/4 + 120/4

y = 135/4

y = 33.75

Therefore, the height of the power line, when it is 30 feet away from one of the poles, is 33.75 feet. This is the y coordinate of point K.

Side note: 33.75 ft = 33 ft + 0.75 ft = 33 ft + 12*0.75 in = 33 ft + 9 inches

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2 years ago
Draw and identify the coordinates of the image of the figure after a 180° rotation
sladkih [1.3K]

Answer:

Step-by-step explanation:

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Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )
Zina [86]

Answer:

\purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

Step-by-step explanation:

\cos( \alpha ) =  \frac{ \sqrt{6} }{6}  =  \frac{1}{ \sqrt{6} }  \\  \\  \therefore \:  \sin( \alpha )  =  \sqrt{1 -  { \cos}^{2} ( \alpha ) }  \\  \\  =  \sqrt{1 -  \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} }  \\  \\ =  \sqrt{1 -  {\frac{1}{ {6} }}}  \\  \\ =  \sqrt{ {\frac{6 - 1}{ {6} }}}   \\  \\  \red{\sin( \alpha ) =  \sqrt{ { \frac{5}{ {6} }}} } \\  \\  \tan( \alpha ) =  \frac{\sin( \alpha ) }{\cos( \alpha ) }  =  \sqrt{5}  \\  \\ \sin( \beta )  =  \frac{ \sqrt{2} }{4}  \\  \\  \implies \: \cos( \beta )  =   \sqrt{ \frac{7}{8} }  \\  \\ \tan( \beta )  =  \frac{\sin( \beta ) }{\cos( \beta ) } =  \frac{1}{ \sqrt{7} }   \\  \\  \tan( \alpha  -  \beta ) =  \frac{ \tan \alpha  -  \tan \beta }{1 +  \tan \alpha .  \tan \beta}  \\  \\  =  \frac{ \sqrt{5} -  \frac{1}{ \sqrt{7} }  }{1 +  \sqrt{5} . \frac{1}{ \sqrt{7} } }  \\  \\  =  \frac{ \sqrt{35} - 1 }{ \sqrt{7}  +  \sqrt{5} }  \\  \\  \purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

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3 years ago
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