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2 years ago

Which equation does the graph of the system of equations solve?

Mathematics

1 answer:

jasenka [17]2 years ago

3 0

Answer:

Step-by-step explanation:

Equation of line 1:

Choose two points : (-1, 0) & (0,2)

y -intercept = b = 2

y = mx+ 2

Plugin the values of the points ( -1 , 0) in the above equation

0 = -1m + 2

-2 = -m

m = 2

Equation of line 1 : y = 2x + 2

Equation of line 2:

(5,0) & (0,5)

y-intercept = b = 5

y = mx +b

y = mx + 5

Plugin the value of points (5 , 0) in the above equation

0 = 5m + 5

-5 = 5m

-5/5 = m

m = -1

Equation of line 2: y = -x + 5

Conclusion: 2x + 2 = -x + 5

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PLEASE HELPPPPPP!!!!! You are performing an experiment to determine how well plants grow under different light sources. Of the 3

PIT_PIT [208]

The probability that a plant in the experiment receives visible or ultraviolet light is 7/10.

Step-by-step explanation:

The total number of plants in the experiment = 30 plants

Plants that receive visible light alone = 12 plants

Plants that receive ultraviolet light alone = 15 plants

Plants that receive both visible and ultraviolet light = 6 plants

step 1 :

Probability = No. of required events / Total number

step 2 :

P(visible) = 12 / 30

P(ultraviolet) = 15 / 30

P(both) = 6 / 30

step 3 :

P(visible or ultraviolet) = P(visible) + P(ultraviolet) - P(both)

= 12/30 + 15/30 - 6/30

= 21/30

P(visible or ultraviolet) = 7/10

8 0

3 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

3 years ago

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Nutka1998 [239]

Answer:

A. W

Step-by-step explanation:

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8 0

3 years ago

What steps should be followed to solve the inequality -4x < 8.

marishachu [46]

Answer:

The require step to solve the inequality

$-4x < 8$ is