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2 years ago

Please help me asap

Mathematics

1 answer:

topjm [15]2 years ago

3 0

To subtract fractions, we need an lcd( lowest common denominator)

8. the lcd is 42 because it’s the smallest number that 6 and 7 are both factors of. the lcd is the bottom of the fraction

to make the denominator of (4/7) 42, we will multiply the top and bottom by 6 and we get (24/42)

to make the denominator of (3/6) 42, we will multiply the top and bottom by 7 and we get (21/42)

these two numbers now have the same denominator so we subtract them

24/42 - 21/42 = 3/42 which simplifies to 1/14

9. since these fractions already have a common denominator, we will make them mixed fractions

-5(2/7) as a mixed fraction. first multiply 5*7 which is 35. now add 2 which is 37. our mixed fraction is -37/7

-1(5/7) as a mixed fraction. multiple 1*7 which is 7. now add 5 which is 12. our mixed fraction is -12/7

now subtract (-37/7) - (12/7) and we get -49/7 which simplifies to -7

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What is the coefficient in the expression 10x + 8 I need help

koban [17]

Answer:

Step-by-step explanation:

A coefficient is a multiplicative factor. Whatever is in front of the variable is a coefficient

5 0

3 years ago

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

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3 years ago

Which is an equation for the line with a y -intercept of zero and a slope of 5?

Vitek1552 [10]

Answer:

y = 5x

Step-by-step explanation:

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3 years ago

What is the perimeter?<br> Help plz....And no links!! I repeat No links!!!

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Step-by-step explanation:

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3 years ago

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4. Find the value of x & y from the following equation 4x+5y=90, x+y=20 a. 10, 15 b. 10, 5 c. 10, 8 d. 10, 10

ira [324]

An equation is formed of two equal expressions. In the two of the given equations, the value of x and y are 10 and 10, respectively. The correct option is D.

<h3>What is an equation?</h3>

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

Given the two of the equation, which can be named as,

4x+5y=90 ...... equation 1

x+y=20 ............ equation 2

Solve the second equation for x,

x + y = 20

x = 20 - y ................ equation 3

In the first equation substitute the value of x from the third equation,

4x + 5y = 90

4(20 - y) + 5y = 90

Solving the equation for y,

80 - 4y + 5y = 90

y = 90 - 80

y = 10

Substitute the value of y in the second equation, to get the value of x,

x + y = 20

x + 10 = 20

x = 20 - 10

x = 10

Hence, In the two of the given equations, the value of x and y are 10 and 10, respectively.

Learn more about Equation here:

brainly.com/question/10413253

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3 0

1 year ago

Read 2 more answers

Please help me asap​

Please help me asap