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2 years ago

5

which equation represents the line that passes through the point (-2,2) and is parallel to y=1/2x + 8

1 answer:

Nataliya [291]2 years ago

5 0

The first thing is the slope
y = 1/2 x + b is the new line. Now we must solve for b
2 = 1/2 (-2) + b
2 = -1 + b
3 = b
The new line is
y = 1/2 x +3

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Verify a ÷ b ≠ b ÷ a , if a = −3/4 and b = 4/5

Shalnov [3]

Given:

$a=-\dfrac{3}{4}$

$b=\dfrac{4}{5}$

To verify:

$a\div b\neq b\div a$

Solution:

We have,

$a=-\dfrac{3}{4}$

$b=\dfrac{4}{5}$

Now,

$a\div b=\dfrac{a}{b}$

$a\div b=\dfrac{-\dfrac{3}{4}}{\dfrac{4}{5}}$

$a\div b=-\dfrac{3}{4}\times \dfrac{5}{4}$

$a\div b=-\dfrac{15}{16}$ ...(i)

And,

$b\div a=\dfrac{b}{a}$

$b\div a=\dfrac{\dfrac{4}{5}}{-\dfrac{3}{4}}$

$b\div a=\dfrac{4}{5}\times (-\dfrac{4}{3})$

$b\div a=-\dfrac{16}{15}$ ...(ii)

From (i) and (ii), it is clear that

$a\div b\neq b\div a$

Hence verified.

5 0

2 years ago

2c – 4 = 3(4 – 2c)<br><br> please find c

astraxan [27]

Answer:

c = 2

Step-by-step explanation:

2c – 4 = 3(4 – 2c)

Expand brackets:

⇒ 2c – 4 = 12 - 6c

Add 6c to both sides:

⇒ 8c – 4 = 12

Add 4 to both sides:

⇒ 8c = 16

Divide both sides by 8:

⇒ c = 2

6 0

2 years ago

Read 2 more answers

Do you use 8 cups of flour to make 24 muffins at that rate how much flour will it take to make 30 muffins

VladimirAG [237]

10 because you do 8/24 then that answer multiplied by 30 to get 10

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2 years ago

When solving an equation in the form by taking square roots, how many solutions will there be?

ss7ja [257]

Normally two. However, one is often extraneous.

3 0

3 years ago

Read 2 more answers

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago