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3 years ago

3/8 plus 1/4 simplest form

Mathematics

2 answers:

marissa [1.9K]3 years ago

8 0

5/8. That is its simplest form.

Artist 52 [7]3 years ago

8 0

In the simplest form it would be 5/8

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Brainly the polynomial x 3 + 5x 2 - 57x - 4189 expresses the volume, in cubic inches, of a shipping box, and the width is (x+3

Simora [160]

Height would be 21 inches, depth would be 5.

8 0

3 years ago

Suppose that 20% of adults belong to health clubs, and 51% of these health club members go to the club at least twice a week. Wh

Westkost [7]

Answer:

10.2% of adults will belong to health clubs and will go to the club at least twice a week

Step-by-step explanation:

assuming that the event H=an adult belongs to a health club and the event T= he/she goes at least twice a week , then if both are independent of each other:

P(T∩H)= P(H)*P(T) ( probability of the union of independent events → multiplication rule )

replacing values

P(T∩H)= P(H)*P(T) = 0.20 * 0.51 =0.102

then 10.2% of adults will belong to health clubs and will go to the club at least twice a week

5 0

3 years ago

Do anybody play valorant?

BartSMP [9]

Answer:

i do, but i just started.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

) 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

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

5 0

3 years ago

The United States government used to make coins of many different values

Papessa [141]

Answer:

1. 50%

2. 3%

3. 20%

4. 200%

Step-by-step explanation:

Here, we want to find the worth of each coin as a percentage of $1

We simply divide the worth by $1 and multiply by 100%

$1 is same as 100 cents

Thus;

1. 50 cent

= 50/100 * 100% = 50%

2. 3 cents

= 3/100 * 100% = 3%

3. 20 cents

= 20/100 * 100% = 20%

4. $2

= 2/1 * 100% = 200%

5 0

2 years ago