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3 years ago

2.3 (x + 1.4) = -9.66

Mathematics

2 answers:

Andre45 [30]3 years ago

7 0

Answer:

- 5,6

Step-by-step explanation:

x + 1,4 = - 4,2

x = - 4,2 - 1,4 = - 5,6

Olin [163]3 years ago

6 0

Answer:

X = -5.6

Step-by-step explanation:

2.3 (x + 1.4) = -9.66

2.3x + 3.22 = -9.66

2.3x = -9.66 - 3.22

2.3x = - 12.88

x = -12.88/2.3

x = -5.6

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Why is glucose converted to ATP if they are both energy?

Sauron [17]

Answer:

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Step-by-step explanation:

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3 years ago

Which expression represents the area of the garden in square feet?

Kisachek [45]

Answer:

C. 42x² - 3x - 9

Step-by-step explanation:

Area:

(7x + 3) (6x - 3)

7x(6x - 3) + 3(6x - 3)

42x² - 21x + 18x - 9

42x² - 3x - 9

8 0

3 years ago

The number of bagels sold daily for two bakeries is shown in the table.

bonufazy [111]

The correct statement based on the data is: A. Mean, because there are no outliers to affect the data.

<h3>What are Outliers?</h3>

Outliers are data point that appears extreme from other data values of a data distribution. When there is an outlier in a data distribution, the median is a better measure of center, while the mean is best when outliers are absent.

From the data distribution for both bakeries, no data seem extreme from the rest of the data, therefore, the mean is a better measure of center.

The answer is: A. Mean, because there are no outliers to affect the data.

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4 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

3 years ago

If w:8=15:24.the value of w is?

luda_lava [24]

Answer:

Step-by-step explanation:

24 divided by 3 is 8

so 15 divided by 3 is 5

3 0

3 years ago