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aev [14]
2 years ago
15

Determine if the given equations are parallel, perpendicular, or neither 4y=3x-4 and 4x+3y=-6

Mathematics
1 answer:
Kazeer [188]2 years ago
8 0

Answer:

Perpindicular

Step-by-step explanation:

Solve each equation for y, or point Slope form.

y = mx + b

4y = 3x - 4 divide both sides by 4 to isolate the y.

y = 3/4x - 1

The second equation.

4x + 3y = - 6 subtract 4x from both sides

3y = -4x - 6 divide both sides by 3

y = -4/3 - 2

If you visualize line 1 having a Slope of 3/4, that is, for every 4 we move in the positive x direction, y will increase by 3, it slopes up from left to right crossing the y axis at (0,-1). The second line moves 3 in the positive x direction and decreases in y by 4, it slopes down from left to right crossing the y axis at (0,-2}.

Because the two lines have negative inverse slopes they are perpendicular. 3/4x and - 4/3x

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Answer:

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Step-by-step explanation:

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3 years ago
7th grade 30 points " How many stage directors are in the play "A christmas carol"
Nadusha1986 [10]

Answer: the answer is e:5

Step-by-step explanation:

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3 years ago
Need help ASAP! Choose the numbers that are irrational. 1. 0. 87 repeating 2. 3.624 3. √2 4. 0. 6 repeating 5. -5 6. 1 over 3 7.
svp [43]

Answer:

√2, pi.

Step-by-step explanation:

An irrational number cannot be written as a fraction a/b where a and b are integers ( not = 0)..

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2 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
What is the value of a^2 + 3b + c - 2d, when a = 3, b = 8, c = 2, and d = 5?
alekssr [168]

Answer:

25

Step-by-step explanation:

'. a=3

b=8

c =2

d=5

so

3^2+3(8)+2-2(5).

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=25

8 0
3 years ago
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