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vodomira [7]
3 years ago
9

Find the distance between (3,-2) and (2,-4)

Mathematics
1 answer:
Sphinxa [80]3 years ago
4 0

Answer:

i believe it is (2,1)

Step-by-step explanation:

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Texas airlines lose 32,400 passengers last week the airline use only airplanes that have a capacity of 216 passengers which equa
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I hope this helps u
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Which of the following are written in point-slope
tino4ka555 [31]
If I get a thanks I’ll answer on a diff comment
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13 points plus brainliest to the correct answer!
TEA [102]
The equation of a straight line is given by:

y = mx + c

where m is the slope and c is the y-intercept.

From the given graph, the y-intercept is 1.

The slope of a straight line is given by:

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

Where (x_1,y_1) and (x_2,y_2) are two points on the line.

From the graph, (0, 1) and (14, 8) are two points on the line, thus:

m= \frac{8-1}{14-0} = \frac{7}{14} = \frac{1}{2}

Therefore, the equation representing the graph is y= \frac{1}{2} x+1
8 0
3 years ago
Read 2 more answers
143=7+8(3-7x) solve and justify
Llana [10]
Justify means plug answer in and verify

so distribute 8(3-7x)=24-56x

143=7+24-56x
143=31-56x
minus 31 both sides
112=-56x
divide both sides by -56
-2=x

justify

plug it back
143=7+8(3-7x)
143=7+8(3-7(-2))
143=7+8(3-(-14))
143=7+8(3+14)
143=7+8(17)
143=7+136
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true

x=-2

7 0
3 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
astra-53 [7]

Answer:

3\pi \rightarrow y=2\cos \dfrac{2x}{3}\\ \\\dfrac{2\pi }{3}\rightarrow y=6\sin 3x\\ \\\dfrac{\pi }{3}\rightarrow  y=-3\tan 3x\\ \\10\pi \rightarrow y=-\dfrac{2}{3}\sec \dfrac{x}{5}

Step-by-step explanation:

The period of the functions y=a\cos(bx+c) , y=a\sin(bx+c), y=a\sec (bx+c) or y=a\csc(bx+c) can be calculated as

T=\dfrac{2\pi}{b}

The period of the functions y=a\tan(bx+c) or y=a\cot(bx+c) can be calculated as

T=\dfrac{\pi}{b}

A. The period of the function y=-3\tan 3x is

T=\dfrac{\pi}{3}

B. The period of the function y=6\sin 3x is

T=\dfrac{2\pi}{3}

C. The period of the function y=-4\cot \dfrac{x}{4} is

T=\dfrac{\pi}{\frac{1}{4}}=4\pi

D. The period of the function y=2\cos \dfrac{2x}{3} is

T=\dfrac{2\pi}{\frac{2}{3}}=3\pi

E. The period of the function y=-\dfrac{2}{3}\sec \dfrac{x}{5} is

T=\dfrac{2\pi}{\frac{1}{5}}=10\pi

5 0
3 years ago
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