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4 years ago

in 4.5 hours, 11.25 inches of snow fell suppose the same amount of snow fell each hour how much snow fell each hour

1 answer:

6 0

If 11.25 inches snow fell in 4.5hrs,
Then for each hr how much snow fell?

So,
4.5 : 11.25
To find how many inches that fell per hour who would need to divide the number of inches by the numbers of hours

11.25 divided by 4.5 = 2.5

So the answer would be 2.5 inch was of snow per hour

You might be interested in

Find the value of x.

kirza4 [7]

Hey there :)

Here we need to use Trigonometry!

The given angle is 61°
The opposite to the angle is 22
So x is the adjacent

The trigonometric ratio that uses opposite and adjacent is tan
( I have uploaded a picture to help you )

tan α = $\frac{opposite}{adjacent}$
tan 61 = $\frac{22}{adj}$
adj = $\frac{22}{tan(61)}$
adj ≈ 12.19 ≈ 12

x = 12

5 0

3 years ago

Rose bought some spinach. She used 3/5 of the spinach on a pan of spinach pie for a party and 3/4 of the remaining spinach for a

SashulF [63]

a. She used 1/10 of the spinach for a salad.

b. She used 1/2 pound of spinach for the salad.

We can solve each part of this problem by using a "fraction strip". The strip helps show a breakdown of what portion of spinach was used for each the party, family, and salad. See the attached image for a step by step example of how to multiply the fractions to find the correct portions.

4 0

4 years ago

Read 2 more answers

A truck can travel 80 miles in the same time that ot takes a car to travel 90 miles. If the truck’s rate is 5 miles per hour slo

SIZIF [17.4K]

The truck was going 40 miles per hour

The amount of time was 2 hours:
(90-80: 10)(10/2=5)

80 divided by two is 40

The car was going 45 miles per hour

(90-80:10)(10/2=5)

90 divided by two is 45

Hopefully you find this helpful

3 0

3 years ago

) Which statement describes the expression shown above? 6 times the sum of 2 and 8 ) plus the sum of 6 and 8 ) 8 plus the produc

ycow [4]

Answer: my childhood Memories

We're going on a trip in our favorite rocket ship

Zooming through the sky, Little Einsteins

the black/hood verse

"We're going on to the club in our favorite range rover

Zooming through the hood, Little gangsta's"

4 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

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

5 0

3 years ago