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tamaranim1 [39]

2 years ago

14

Can someone please help me with this

1 answer:

inysia [295]2 years ago

6 0

-2/2 is your slope, down 2, right 2.

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PLZ HELP ill GIVE BRAINIESt!!!!20 Points

Semmy [17]

HOPE YOU CAN READ THE EXCEL FILE

4 0

3 years ago

Soda A cost $1.49 per 28 oz. bottle vs. $0.95 per 21 oz.

IrinaVladis [17]

Answer:

You will be spending for Soda A for about 1 more cent than Soda B

Step-by-step explanation:

Find the rate of both sodas.

Soda A.

1.49 / 28 = $0.0532 per oz.

Soda B.

0.95 / 21 = $0.0452 per oz.

7 0

3 years ago

Caroline knows the height and the required volume of a cone-shaped vase she’s designing. Which formula can she use to determine

timama [110]

Answer:

Option B. $r=\sqrt{\frac{3V}{\pi h}}$

Step-by-step explanation:

we know that

The volume of a cone is equal to

$V=\frac{1}{3}\pi r^{2} h$

Solve for the radius r

That means-----> isolate the variable r

Multiply by 3 both sides

$3V=\pi r^{2} h$

Divide by $(\pi h)$ both sides

$r^{2}=\frac{3V}{\pi h}$

square root both sides

$r=\sqrt{\frac{3V}{\pi h}}$

4 0

3 years ago

Make sure to check the final answer!

LenKa [72]

Answer:

k = 33

Step-by-step explanation:

33 ÷ -11 = -3

SIMPLE DIVISION lol

5 0

2 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