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2 years ago

What is the slope of the line passing through the points (-3, 4) and (2, - 1)?

Mathematics

1 answer:

Tcecarenko [31]2 years ago

5 0

Answer:

Slope = -1

Step-by-step explanation:

m = y^2 - y^1 / x2 - x1

let (x1, y1) = (-3,4) & (x2, y2) = (2,-1)

m = -1-4 / 2-(-3) = -5/5 = -1

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Convert the fraction 5/8 to decimal using long division method

erma4kov [3.2K]

1/8=0.125
0.625*5=0.625
Or
1/2=0.5
0.5/2=1.25
1.25/2=0.125
Which is 1/8

7 0

2 years ago

The 12th term of a sequence is 41 and the 15th term is 140. What is the nth term?

OLga [1]

An= mth term.
an=a₁+(n-1)*d

a₁₂=41
a₁₅=140

a₁₂=41
41=a₁+(12-1)*d
41=a₁+11d
a₁+11d=41 (1)

a₁₅=140
140=a₁+(15-1)*d
140=a₁+14d
a₁+14d=140 (2)

With the equiations (1) and (2) build a system of equations
a₁+11d=41
a₁+14d=140

we solve it.
-(a₁+11d=41)
a₁+14d=140
--------------------
3d=99 ⇒d=99/3=33

a₁+11d=41
a₁+(11*33)=41
a₁+363=41
a₁=41-363=-322

an=a₁+(n-1)*d
an=-322+(n-1)*33
an=-322+33n-33
an=-355+33n

an=-355+33n

To check:
a₁₂=-355+33*12=-355+396=41
a₁₅=-355+33*15=-355+495=140.

3 0

2 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

2 years ago

What is the slope of a line perpendicular to line 6x + 3y = 15

NISA [10]

Step-by-step explanation:

Given line is 6x+3y=15

Slope of the given line (m)=-x/y

=-6/3

=-2

let slope of the line perpendicular to the given line be m'.

Condition of perpendicularity,

m×m'=-1

-2×m'=-1

m'=-1/-2

m'=1/2

8 0

2 years ago

Find out about tall buildings where you live. What is the tallest building in your hometown or state? What is the difference in

Aleks04 [339]

The tallest building in my area is the reunion tower which is 562 ft and the difference is 1106

6 0

2 years ago