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2 years ago

A. x = 2, y =6 B. x = 2, y =3 C. x =-2, y =3 D. x = -3, y =2

Mathematics

1 answer:

Vsevolod [243]2 years ago

7 0

Answer:

the answer is

A. X=2, y=6

Step-by-step explanation:

use substitution

y= 6 so,

y-3 = 6-3 which is 3

and y-1 = 6-1 which is 5

now for x

x=2

x+3= 2+3 which is 5

2x-1= 2(2)-1

2 times 2 is 4

4-1 = 3

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Question below in image

saul85 [17]

33 is correct good job

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2 years ago

Ava and Kelly ran a road race, starting from the same place at the same time. Ava ran at an average speed of 6 miles per hour. K

zalisa [80]

<h2>Hello!</h2>

The answer is:

It will took 0.375 hours for Ava and Kelly to be 3/4 miles apart.

To calculate how long after they start they will be 3/4 miles apart, we need to write two equations.

So, writing the equations, we have:

Calculations for Ava:

We have the following information,

$v_{Ava}=6mph$

Then, writing the equation,

$x_{Ava}=x_{o}+v_{Ava}*t$

$x_{Ava}=x_{o}+v_{6mph}*t$

Calculations for Kelly:

We have the following information,

$v_{Kelly}=8mph$

We need to calculate when Kelly will be 3/4 miles apart of Ava, so, it's position will be the Ava's position plus 3/4 miles.

Then, writing the equation,

$x_{Ava}+0.75miles=x_{o}+v_{Kelly}*t$

$x_{Ava}+0.75miles=x_{o}+v_{8mph}*t$

Now, substituting Ava's speed into the second equation, we have:

$x_{o}+6mph*t+0.75miles=x_{o}+8mph*t$

$6mph*t+0.75miles=+8mph*t$

$8mph*t-6mph*t=0.75miles$

$2mph*t=0.75miles$

$t=\frac{0.75miles}{2mph}=0.375hours$

Hence, we have that it will took 0.375 hours for Ava and Kelly to be 3/4 miles apart.

Have a nice day!

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3 years ago

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figure abcd is similar to figure mnkl if ab = 2x + 1, bc= 12, mn = 8, nk=7 and kl=13, what is the value of x

Kruka [31]

Step-by-step explanation:

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3 years ago

A 6-ounce bottle of juice costs $.96. An 8-ounce bottle costs $1.12. Which is the better to buy?

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3 years ago

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Plutonium-240 decays according to the function L where o

PIT_PIT [208]

Answer:

$\boxed{\textbf{1700 yr}}$

Step-by-step explanation:

$-\dfrac{\text{d}L}{\text{d}t} = kL\\\\\dfrac{dL}{L}=-kdt\\\\\ln L = -kt + C\\\text{At t = 0, L = L$_{0}$, so C = $\ln L_{0}$}\\\ln L = -kt + \ln L_{0}\\\ln L_{0} - \ln L = kt\\\\\ln \dfrac{L_{0}}{L} =kt$

Data:

L₀ = 24 g

L = 20 g

k = 0.000 11 yr⁻¹

Calculation:

$\ln \dfrac{24}{20} =0.000 11t\\\\\ln 1.2 = 0.000 11t\\\\0.1823 = 0.000 11t\\\\t = \dfrac{0.1823}{0.000 11} = \textbf{1700 yr}\\\\\text{It will take } \boxed{\textbf{1700 yr}}\text{ for the polonium to decay to 20 g}$

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3 years ago