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3 years ago

Which of the following points is a solution to the system of equations shown?

1 answer:

5 0

Using Substitute method:

Replace y value of equation 2 into equation 1:
4x + (x+6) = 1

Solve for x:
4x + x + 6 = 1
5x = 1 - 6
5x = -5
x = -5/5
x = -1

Pick any of the two equation, replace x with it's value -1 and solve for y:
y = (-1) + 6
y = -1 + 6
y = 5

Solution: (-1, 5)

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4 years ago

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Answer:

May be the second one

Step-by-step explanation:

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3 years ago

Ozzie has a coupon for 5% off any purchase in his favorite bakery. When Ozzie stops to buy some cookies, he discovers that all c

maria [59]

Answer:

Step-by-step explanation:

you need to first do 5.89- [(5.89 * 15)/100]=5.89-0.8835=5.0065

then 5.0065 * 2 = 10.013 because it is 15% each dozen then

[(10.013 * 5)/100] =0.50065 then 10.013 - 0.50065 = 9.51235

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3 years ago

Read 2 more answers

How many steps do we take from 0 to -13???

Rus_ich [418]

Answer:

Step-by-step explanation:

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3 years ago

Air pressure may be represented as a function of height (in meters) above the

denis-greek [22]

Answer: OPTION C.

Step-by-step explanation:

<h3> The complete exercise is: "Air pressure may be represented as a function of height (in meters) above the surface of the Earth, as shown below:</h3><h3> $P(h) =P_0e^{-0.00012h}$ </h3><h3> In this function $P_o$ is the air pressure at the surface of the earth, and $h$ is the height above the surface of the Earth, measured in meters. At what height will the air pressure equal 50% of the air pressure at the surface of the Earth"</h3><h3></h3>

Given the following function:

$P(h) =P_0e^{-0.00012h}$

In order to calculate at what height the air pressure will be equal 50% of the air pressure at the surface of the Earth, you can follow these steps:

1. You need to substitute $P(h)=0.5P_o$ into the function:

$0.5P_o=P_0e^{-0.00012h}$

2. Finally, you must solve for $h$ .

Remember the following property of logarithms:

$ln(b)^a=a*ln(b)\\\\ln(e)=1$

Then, you get this result:

$0.5P_o=\frac{P_o}{e^{0.00012h}}\\\\(0.5P_o)(e^{0.00012h})=P_o\\\\e^{0.00012h}=\frac{P_o}{0.5P_o}\\\\ln(e)^{0.00012h}=ln(2)\\\\0.00012h*1=ln(2)\\\\h=\frac{ln(2)}{0.00012}\\\\h=5576.2$

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4 years ago