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3 years ago

What would be the first step to slove the system of equations 2x-9y=6 and 4x-3y=-7

2 answers:

5 0

If there are choices, then you need to choose the correct choice, but if there are no choices, then one correct answer is to multiply the first equation by -2 and add it to the second equation to eliminate x and solve for y.

andrew11 [14]3 years ago

4 0

The first step varies depends on what method you want to use to solve the system of equations.
If it is me, my first step would be either doubling the first equation so the coefficients of x in both equations equal to each other, or tripling the second equation so the coefficient of y in both equations is set equal.

2x-9y=6
4x-3y=-7

3(second euation) =12x-9y=-7

second equation subtract the first equation
10x=-13
x=-13/10

Then we can plug x in any of the two original equations and solve for y which i got to be -19/15

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If EF = x + 5, DC = 2x + 4, and AB = 2x + 2. Find EF.<br><br>

bonufazy [111]

Answer:

B *second one down.

Step-by-step explanation:

EF = 1/2(AB + CD)

x+5 = 1/2 (2x + 2 + 2x+4) Combine the like terms on the right.

x+5 = 1/2 (4x + 6) Remove the brackets.

x+5 = 1/2*4x + 1/2*6

x+5 = 2x + 3 Subtract x from both sides

5 = 2x-x + 3

5 = x + 3 Subtract 3 from both sides

5-3 = x

x = 2

EF = x + 5

EF = 2 + 5

EF = 7

The answer is B

3 0

2 years ago

Find the value of x that makes m||n

svet-max [94.6K]

7x-23+11x+23=180
4x+0=180
4x/4=180/4
X=45

Answer: x=45

5 0

3 years ago

If g(x) = 2x + 4, find g(1).

JulijaS [17]

G(1) = 2(1) + 4 = 6
g(1) = 6

6 0

3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

2 years ago

Calculator

gregori [183]

Answer:

9165.0

Step-by-step explanation:

13*94*7.5 = 9165.0

7 0

3 years ago