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Vikentia [17]
3 years ago
9

Find the equation of a line that passes through (1,12) and is parallel to the graph of y=3x+3 Write the equation in​ slope-inter

cept form, if possible.
Mathematics
1 answer:
Ray Of Light [21]3 years ago
5 0

Answer:

The equation of the parallel line is

y = 3x + 9

Step-by-step explanation:

The general equation of a straight line is given as;

y = mx + c

where m is the slope and c is the intercept

If two lines are parallel, their slopes are equal

So the slope of the new line too is 3

Using the point-slope formula

y-y1 = m(x-x1)

y-12 = 3(x-1)

y-12 = 3x-3

y = 3x-3 + 12

y = 3x + 9

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Answer:

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3 years ago
Suppose that p ¼ P(male birth) ¼ .5. A couple wishes to have exactly two female children in their family. They will have childre
BlackZzzverrR [31]

Answer:  the probability that the family has x male children can be found in the explanation.

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3 years ago
what is the most appropriate center for 27 142 145 146 154 162 /12592106/70400e23?utm_source=registration
Klio2033 [76]
<span>27 142 145 146 154 162    

27 is definitely an outlier.  You could temporarily drop it and focus on
</span>{142 145 146 154 162}.  The median of this 5-numeral set is 146; it'd be a good "center."  So would the mean  (<span>142 +145 +146 +154 +162) /5.  No one value shows up more than once, so forget about "mode."

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6 0
3 years ago
If sinx = p and cosx = 4, work out the following forms :<br><br><br>​
Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16} $

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

\boxed{\sin(-x)=-\sin(x)}

\boxed{\cos(-x)=\cos(x)}

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

\sin(x)=p

\cos(x)=4

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]

Using the sum and difference identities:

\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}

\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}

\sin(\pi -x) = \sin(x)

\sin(2\pi +x)=\sin(x)

\cos(2\pi-x)=\cos(x)

Therefore,

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]

\implies [p+4] \cdot [p \cdot 4]=4p^2+16p

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16} $

8 0
3 years ago
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loris [4]
It would have to be B, because with a vertex on the x-axis, the only point it would touch the x-axis is the vertex, thus proving that this quadratic has only one zero.
6 0
3 years ago
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