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3 years ago

Find the equation of the line parallel to the line y = 4x – 2 that passes through the point (1, 5).

Mathematics

1 answer:

Lesechka [4]3 years ago

4 0

Answer:

4x-y+1=0

Step-by-step explanation:

here,given equation of a line id

4x-y-2=0.. eqn(i)

equation of any line parallel to line (i) is

4x-y+k=0...eqn(ii)

since, the line(ii) passes through (1,5)[replacing x=1 and y=5 in eqn(ii), we get]

4*1-5+k=0

or, 4-5+k=0

or,-1+k=0

•°•k=1

substituting the value of k=1 in eqn(ii),

4x-y+1=0 is the required equation of the line.

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1.. A train travels at a speed of 58 km/hr. If it goes from one station to the next in 1.6 hours, how far has it traveled?

wel

Answer:

92.8 km

Step-by-step explanation:

s=d/t

58=d/1.6

58*1.6=d

92.8=d

7 0

3 years ago

What's the probability of pulling two red cards in a row from a deck of cards?

monitta

Hey there! I'm happy to help!

Let's say that we don't have jokers. In that case, there are 52 cards, and half of them are red (so 26 are red). The probability of pulling a red card once is 1/2 since half the cards are red.

If we pick one out, there are only 51 total cards left and 25 red cards. So, the probability of picking one again would be 25/51.

We multiply the probabilities of these two events to find the probability of them both happening.

1/2×25/51=25/102

The probability of picking two red cards in a row is 25/102 or around 24.5%.

Have a wonderful day! :D

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3 years ago

A rectangle has a width of 9 cm, and a diagonal of 15 cm. Find the length.

iren2701 [21]

The answer is
$9 { + 15 {}^{2} }^{= \sqrt{108}$

6 0

3 years ago

Someone please help my with my geometry test?!!!

jenyasd209 [6]

Answer:

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2 years ago

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

Zigmanuir [339]

Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

4 0

3 years ago