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3 years ago

Solve the equation 1.6=7.6-5(k+1.1)

1 answer:

8 0

7.6 - 5(k + 1.1) = 1.6

7.6 - 5(k + 1.1) - 7.6 = 1.6 - 7.6

-5(k +1.1) = -6

-5(k + 1.1) -6
-------------- = ---------
-5 -5

k + 1.1 = 6
-----
5

k + 1.1 - 1.1 = 6
----- - 1.1
5

k = 0.1

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Please someone tell me the answers to these I dont remember how to do it

Delvig [45]

Answer:

1) (3 - 6x)(-8) = (-8)(3) - (-8)(6x) = -24 - (-48x) = -24 + 48x = 48x - 24

2) (-12)(2x - 3) = (-12)(2x) - (-12)(3) = -24x - (-36) = -24x + 36 = 36 - 24x

3) (10 - 2x)9 = (9)(10) - (9)(2x) = 90 - 18x

4) (-5)(11x - 2) = (-5)(11x) - (-5)(2) = -55x - (-10) = -55x + 10 = 10 - 55x

5) (1 - 9x)(-10) = (-10)(1) - (-10)(9x) = -10 - (-90x) = -10 + 90x = 90x - 10

6) (-6)(x + 8) = (-6)(x) + (-6)(8) = -6x + (-48) = -6x - 48

7) (-4 + 3x)(-8) = (-8)(-4) + (-8)(3x) = 32 + (-24x) = 32 - 24x

8) (-5)(1 - 11x) = (-5)(1) - (-5)(11x) = -5 - (-55x) = -5 + 55x = 55x -5

9) (-12x + 14)(-5) = (-5)(-12x) + (-5)(14) = 60x + (-70) = 60x - 70

Step-by-step explanation:

The distributive property is a(b + c) = ab + ac

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2 years ago

Which of the following points satisfies the inequality y>1/3x-2 which side should be shaded

garik1379 [7]

Answer: Hope this helps <3

Step-by-step explanation:

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

Please I need the polynomial for this problem, if you don’t know then don’t answer! Wrong answers will be reported

babymother [125]

Answer:

Area = (36π - 12πx + x²π) mm²

Step-by-step explanation:

Given that,

radius of lense = 6mm

Length from outer layer of pupil to lense = x mm

Radius of pupil = 6 - x mm

<h3>Formula to calculate area of pupil</h3><h3> Area = π r²</h3>

where r is radius of pupil

Area = π (6 -x)²

Area = π (6² -2(6)(x) + x²)

Area = π (36 -12x +x²)

Area = (36π - 12πx + x²π) mm²

So, the polynomial in standard form that represent the area of pupil in terms of π = (36π - 12πx + x²π) mm²

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3 years ago

What is the volume of a rectangular prism with a length of 2.5 cm, a width of 3.1 cm and a height of 1.2 cm?

Svetradugi [14.3K]

The answer is D- 9.3cm^3

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3 years ago