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3 years ago

Find an equation of the line that passes through the point (23, 22) and is parallel to 2x=3-5y

Mathematics

2 answers:

Naya [18.7K]3 years ago

8 0

Y=mx+b

Slope is 3/1

Y-22=3(x-23)
Y-22=3x-69
+22 +22
Y=3x -47

Y=3x-47

Korvikt [17]3 years ago

5 0

Answer:

The equation of the line is y= -2/5x+156/5

Step-by-step explanation:

y=mx+b

m= -2/5 ( 23,22)

y-y=m(x-x¹)

y-22=-2/5(x-23)

y-22=-2/5x+46/5

+22 +22

y= -2/5x+156/5

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iris [78.8K]

Answer:

Option 4

Step-by-step explanation:

Term = n/(4^n)

(n+1)th term = (n+1)/[4^(n+1)]

Ratio: (n+1)/[4^(n+1)] ÷ n/(4^n)

= (n+1)/[4^(n+1)] × (4^n)/n

= [(n+1)/n] × (1/4)

For large n, (n+1)/n approaches 1

[(n+1)/n] × (1/4) approaches 1/4 which is less than 1

Since the ratio is less than 1, series converges

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maks197457 [2]

Backing out the bonus, we get 2711 - 250 = 2461, which is 161 more than his last check

<span>So 161 / 2300 = 0.07, or 7%</span>

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3 years ago

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Virty [35]

Answer: The mean is 3780

Step-by-step explanation:

Mean is found by adding all values and dividing the sum by how many values were added.

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3 years ago

nasty-shy [4]

Answer:

Step-by-step explanation:

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2 years ago

The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati

makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

$y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\$

And the second one would be

$y = 2-x \,\,\,\,\,, 1 \leq x \leq 2$ .

If you rotate the first region around the "y" axis you get that

$A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51$

And if you rotate the second region around the "y" axis you get that

$A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188$

And the sum would be 2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first region around the x axis you get that

$A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708$

And if you rotate the second region around the x axis you get that

$A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472$

And the sum would be 1.5708+1.0472 = 2.618

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3 years ago