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3 years ago

Rule: y=x+3 HELPpppppp

Mathematics

2 answers:

dimaraw [331]3 years ago

7 0

Answer:

Step-by-step explanation:

when y=4

x=4-3=1

when y=8

x=8-3=5

when y=5

x=5-3=2

aivan3 [116]3 years ago

4 0

Answer:

$\left[\begin{array}{ccc}x&y\\1&4\\5&8\\2&5\end{array}\right]$

Step-by-step explanation:

x + 3 = y

1 + 3 = 4

5 + 3 = 8

2 + 3 = 5

brainliest?

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Select the correct answer.

mariarad [96]

Answer:

D. slope: 3

y-intercept: -3

Step-by-step explanation:

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2 years ago

Please help will give brainliest

Vladimir [108]

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$\frac{4}{5} + \frac{3}{15} - \frac{2}{3} \\$

Taking LCM

$\frac{12 + 3 - 10}{15} \\ \\ \longrightarrow \: \cancel\frac{5}{15} \\ \\ \longrightarrow \: \frac{1}{3}$

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1 year ago

When an object falls freely, the formula for the distance it falls is d= 4.9t^2 where d is

german

Answer:

a) 122.5 m ; b) 10 s

Step-by-step explanation:

Just use the given equation and plug in what you know.

d = 4.9t^2

d = 4.9(5^2)

d = 4.9 * 25

d = 122.5 m

d = 4.9t^2

490 = 4.9t^2

100 = t^2

t = 10 s

3 0

2 years ago

Read 2 more answers

Which fraction is equal to 1/5. Can someone please answer this question

NeX [460]

Answer:

2/10

20/100

5/25

Step-by-step explanation:

I hope this helps you

4 0

1 year ago

Read 2 more answers

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

2 years ago