Solve the system of equations Y=-2x+5 and y=x^2+3x+9

Katie gives Maggie half of her pencils. Maggie keeps 5 pencils and gives the rest to Jamil. A. Write an expression for the numbe

Alexandra [31]

Here's what I mean when I write x/2 JUST IN CASE.

а.

Let the total number of pencils be x, then this implies that Maggie received x/2 pencils from Katie. Maggie keeps 5 and give the rest to Jamil, this means

that Jamil receives

x/2-5 pencils.

b. Expression of pencils given to Jamil is: x/2-5

Substitute x=16 in this expression: x/2-5

Simplify and evaluate: = 8-5= 3

Maggie gives 3 pencils to Jamil.

c. Expression of pencils given to Jamil is: x/2-5 to

Equate this expression with 1 to form an equation and solve for x all, the total

number of pencils if Jamil received 1 pencil:

x/2 - 5 = 1

Solve for x:

x = 2(1+ 5)

Evaluate:

x = 2(6) = 12

Katie had 12 pencils if Jamil received 1 pencil.

RESULT

a. Expression of number of pencils given to Jamil is x/2-5, where x is total number of pencils.

b. Maggie gives 3 pencils to Jamil.

c. Katie had 12 pencils if Jamil received 1 pencil.

5 0

3 years ago

Divide three fourths ÷ two fifths

KengaRu [80]

Your answer would be on eof the three
15/8
1.875
1 7/8
Hope this helps...

7 0

3 years ago

Read 2 more answers

Consider the initial value problem yâ€²+2y=4t,y(0)=8.

Xelga [282]

Answer:

Please read the complete procedure below:

Step-by-step explanation:

You have the following initial value problem:

$y'+2y=4t\\\\y(0)=8$

a) The algebraic equation obtain by using the Laplace transform is:

$L[y']+2L[y]=4L[t]\\\\L[y']=sY(s)-y(0)\ \ \ \ (1)\\\\L[t]=\frac{1}{s^2}\ \ \ \ \ (2)\\\\$

next, you replace (1) and (2):

$sY(s)-y(0)+2Y(s)=\frac{4}{s^2}\\\\sY(s)+2Y(s)-8=\frac{4}{s^2}$ (this is the algebraic equation)

b)

$sY(s)+2Y(s)-8=\frac{4}{s^2}\\\\Y(s)[s+2]=\frac{4}{s^2}+8\\\\Y(s)=\frac{4+8s^2}{s^2(s+2)}$ (this is the solution for Y(s))

c)

$y(t)=L^{-1}Y(s)=L^{-1}[\frac{4}{s^2(s+2)}+\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+L^{-1}[\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}$

To find the inverse Laplace transform of the first term you use partial fractions:

$\frac{4}{s^2(s+2)}=\frac{-s+2}{s^2}+\frac{1}{s+2}\\\\=(\frac{-1}{s}+\frac{2}{s^2})+\frac{1}{s+2}$

Thus, you have:

$y(t)=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}\\\\y(t)=L^{-1}[\frac{-1}{s}+\frac{2}{s^2}]+L^{-1}[\frac{1}{s+2}]+8e^{-2t}\\\\y(t)=-1+2t+e^{-2t}+8e^{-2t}=-1+2t+9e^{-2t}$