Please help will give brainlest!!

Write the equations for the basic absolute value equation y=|x-h|+k under the given transformations:

BlackZzzverrR [31]

Left & Right affect the h,
Up & Down affect the k,
Shrink & Stretch affect the "a" (which is in front of the absolute value expression).

1. shift 1 unit to the right and up 2 units → y = |x-1| + 2
2. shift 3 units to the left and 7 units down → y = |x + 7| - 7
3. vertical shrink by a factor of 1/3 → y = $\frac{1}{3}$ |x|
4. vertical stretch by a factor of 3 → y = 3 |x|

5 0

3 years ago

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

Step 2:-

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

7. Stan is serving lemonade at the school's dance out of no

Lilit [14]

Answer: 540 cups

Step-by-step explanation:

Diameter = 20

Radius = 20/2=10

Volume of a cylinder: pi(r^2)h = 3600pi

Cone = (pi(r^2)h)/3 = (20pi)/3

3600pi/(20pi/3) = 540 cups

7 0

3 years ago

Read 2 more answers

The sum of two numbers is 18. 5 times the first number subtracted from 4/7 of the second number is 14. Find the number.

Sergeu [11.5K]

Answer:

The sum of two numbers is 18.

5 times the first number subtracted from 4/7 of the second number is 14.

First number --> x

Second number --> y

so

x + y = 18 --> x = 18-y

5x - (4/7)y = 14

so

5 (18-y) - (4/7)y = 14

90 - 5y - 4/7 y = 14

-5.57 y = 14-90

y = 13.64

so

x = 18 - 13.64

x = 4.36

8 0

2 years ago

One way to show that two triangles are similar is to show that

avanturin [10]

Answer:

show that the three sets of corresponding sides are in proportion. If the three sets of corresponding sides of two triangles are in proportion, the triangles are similar.

Step-by-step explanation:

7 0

2 years ago