Which of the following relations is not a function? x y 0 1 1 4 -1 4 x y 0 2 1 3 1 -3 x y 0 -2 1 -3 -2 5 x y 0 -2 1 -2 2 -2

Samantha has a pig enclosure that is 4 feet by 6 feet in her backyard. She wants to double the area of the enclosure by increasi

luda_lava [24]

Answer:

Step-by-step explanation:

The answer is 48 feet because 4x6=24 and 24x2=48

4 0

3 years ago

Anything helps please help me!

dimaraw [331]

5 1/2 = 11/2
11/2 x 3/4 = 33/8 = 4 1/8
answer is B. 4 1/8

8 0

3 years ago

One leg of a triangle is 7 inches. another leg measures 5 inches. if the perimeter of the triangle is 19 inches, find the length

Setler79 [48]

SOLUTION

A triangle has three sides. The sides of this triangle are 7 inches, 5inches and the unknown side.

Let the unknown side be x inches.

Perimeter is distance around a plane shape.

The perimeter of this triangle is 19,

$\begin{gathered} \text{ That is the perimeter 1}9=7+5+x \\ 19\text{ = 12 }+x \\ x\text{ = 19 - 12 } \\ x\text{ = 7 inches } \end{gathered}$

Therefore, the other side is 7 inches

6 0

1 year ago

What is the next term of the geometric sequence?

Alborosie

Answer:

320

Step-by-step explanation:

5, 20, 80,5,20,80,5, comma, 20, comma, 80, comma is not a geometric sequence.

However, if you mean 5, 20, 80 it is a geometric sequence.

It is being multiplyed by 4 each time

so 80*4=320

The next number would be 320.

4 0

3 years ago

(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

Explanation:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

Step(i):-

Given differential problem

y′+3 y=9 t

Take the Laplace transform of both sides of the differential equation

L( y′+3 y) = L(9 t)

Using Formula Transform of derivatives

L(y¹(t)) = s y⁻(s)-y(0)

By using Laplace transform formula

$L(t) = \frac{1}{S^{2} }$

Step(ii):-

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

Step(iii):-

By using partial fractions , we get

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

B = 9/3 = 3

Put s = -3 in equation(i)

9 = C(-3)²

C = 1

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

0 = A + C

put C=1 , becomes A = -1

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

Step(iv):-

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

By using inverse Laplace transform

$L^{-1} (\frac{1}{s} ) =1$

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

Final answer:-

Now the solution , we get

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

5 0

3 years ago