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Reptile [31]
3 years ago
5

When using base-10 blocks to divide 452 by 2, how many hundreds flats will be in each equal group? hundreds flats Divide 452 by

2 using base-10 blocks - 4 hundreds flats, 5 tens rods, and 2 ones cubes placed side-by-side
Mathematics
1 answer:
daser333 [38]3 years ago
8 0
<h3>Given</h3>

4 hundreds flats; 5 tens rods; 2 ones cubes

<h3>Find</h3>

The number of hundreds flats in each of 2 equal piles

<h3>Solution</h3>

When 4 flats are divided into two equal groups, each group will have ...

... 2 flats

_____

You can imagine doing this the way a card dealer might: first put 1 flat in each of 2 piles, then do the same for the remaining 2 flats. Each pile will end up with 2 flats.

— — — — —

You will have a problem if you continue with the tens rods. There is an odd number of those, so one of them will have to be exchanged for 10 ones cubes.

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jeka57 [31]

Answer:

13.68

Step-by-step explanation:

72 * 0.19 = 13.68

19% can be converted to 19/100 or .19 to change out of a percentage. .19 is the multiplied by 72 to find 19% of 72.

For example

a fourth of something is the same as 25%.

if you want to find 1/4 of 8 you would multiply 8 by 1/4.

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1 year ago
Choose the situation in which using exact numbers is the MOST appropriate.
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A, you need to pay with exact change

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An architect is designing a gym for a new elementary
Liono4ka [1.6K]

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Help asap please i don't know the answer lol
Sveta_85 [38]

Answer:

H

Step-by-step explanation:

The parallel line BC divides the sides AD and AE in proportion, that is

\frac{AB}{BD} = \frac{AC}{CE}, that is

\frac{2}{3} = \frac{AC}{4} ( cross- multiply )

3AC = 8 ( divide both sides by 3 )

AC = \frac{8}{3} = 2 \frac{2}{3}, hence

AE = AC + CE

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6 0
4 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
4 years ago
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