Need help , if you can answer it pls do . thank you

Meridith runs at a rate of 190 meters per minute. use the formula d= r times t to find how far she runs in 6 minutes.

Mariulka [41]

D= r×t
D= 190×6
D= 1140 meters

6 0

2 years ago

Jennie charges $200 per passenger for a sunset sailing

Over [174]

Answer:

1692.5 and -232.5

Step-by-step explanation:

6 0

2 years ago

Solve for X Please walk me through the steps, I understand most of it I just forgot how to do it

nadya68 [22]

Answer:

x=8

Step-by-step explanation:

first you need to find the unknown corner so you do 180-71= 109

then find the interior angle degree (n-2)*180 (n=number of sides) so (4-2)*180=2(180)=360

then set everything equal to 360:

10x+6+13x-2+8x-1+109=360

(combine like terms)

31x+112=360

(subtract 112 form both sides)

31x=248

(divide both sides by 31)

x=8

8 0

3 years ago

The polynomial 3x2 is a with a degree of .

ZanzabumX [31]

To determine the degree of a polynomial, you look at every term:

if the term involves only one variable, the degree of that term is the exponent of the variable
if the term involves more than one variable, the degree of that term is the sum of the exponents of the variables.

So, for example, the degree of $7x^{55}$ is 55, while the degree of $xy^2z^4$ is $1+2+4 = 7$

Finally, the term of the degree of the polynomial is the highest degree among its terms.

So, $3x^2$ is a degree 2 polynomial (although it only has one term)

similarly, $x^2y + 3xy^2 + 1$ is a degree 3 polynomial: the first two terms have degree 3, because they have exponents 2 and 1.

6 0

2 years ago

Read 2 more answers

(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

2 years ago