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Romashka-Z-Leto [24]
3 years ago
8

Y=(3/4)x-1 Is it parallel?

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

Answer:

I don't have the answer but I do have an example

Example:

Solution: The slope of the given line, y = -4x + 5 is -4 (remember y = mx + b).

Since parallel lines have = slopes, our line has a slope of -4.

Use the point -slope form of the equation of a line: y - y1 = m(x - x1).

m = -4      (x1, y1) = (6,-3)

ANSWER:    y - (-3) = -4(x - 6)

y + 3 = -4x + 24

y = -4x + 21

Hope this helps

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The diagram below shows a proportional relationship between the number of bags of chips and the price.
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it would be B 3.09 per bag of chips

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Solve the following pair of simultaneous:<br><img src="https://tex.z-dn.net/?f=%205x%20%2B%202y%20%3D%20%20-%202%20%20%20%20%20%
Vladimir [108]

Answer:

x = -2 , y = 4

Step-by-step explanation:

given:

5x + 2y = -2                            <em>*4</em>

4x + 3y = 4                             <em>*5</em>

Solving simultaneously,

    20x + 8y = -8 ....eq 1

(-)  20x +15y = 20  ...eq 2

.............................................

-7y = -28

y = -28 / -7

y = 4

  • if y = 4

using equation 2,

20x +15(4) = 20

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8 0
3 years ago
Use the given information to find the exact value of the trigonometric function
eimsori [14]
\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}

Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

The answer is none of the choices

7 0
1 year ago
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I believe the answer is 127

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