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Reil [10]
3 years ago
13

Annette has finished 20 of her homework problems. That is 80% of the total assinged by her math teacher. How many pronlems were

assinged?
The possible answers are:
16
25
30
32
Mathematics
1 answer:
horrorfan [7]3 years ago
7 0
So 20=80%' % means out of 100 so 80%=80/100 so 20=80/100 we try to get to 100/100 so 20=80/100 5=20/100 5 times 20=100 so 25=100/100 there were 25 questions total
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Can a quadrilateral be a rhombus ?
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Yes it can.  It doesn't <em>have to be</em>, but it can be without too much trouble.

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40 cm = ____ mm what is the answer
Alekssandra [29.7K]
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8 0
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Multiply the polynomials (x+3) (2x-4). What is the product in the form ax2 + bx + c
Angelina_Jolie [31]

Answer:

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Step-by-step explanation:

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Simplify this expression 3xy + 10xy - 5x + 2x - 2yx? Re-write the equations so that like terms are together. (Remember that orde
Elenna [48]
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3 years ago
Solve y" + y = tet, y(0) = 0, y'(0) = 0 using Laplace transforms.
irina1246 [14]

Answer:

The solution of the diferential equation is:

y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}

Step-by-step explanation:

Given y" + y = te^{t}; y(0) = 0 ; y'(0) = 0

We need to use the Laplace transform to solve it.

ℒ[y" + y]=ℒ[te^{t}]

ℒ[y"]+ℒ[y]=ℒ[te^{t}]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]+s·y(0)-y'(0)=s²·Y(s)

ℒ[y]=Y(s)

ℒ[te^{t}]=\frac{1}{(s-1)^{2}}

So, the transformation is equal to:

s²·Y(s)+Y(s)=\frac{1}{(s-1)^{2}}

(s²+1)·Y(s)=\frac{1}{(s-1)^{2}}

Y(s)=\frac{1}{(s^{2}+1)(s-1)^{2}}

To be able to separate in terms, we use the partial fraction method:

\frac{1}{(s^{2}+1)(s-1)^{2}}=\frac{As+B}{s^{2}+1} +\frac{C}{s-1}+\frac{D}{(s-1)^2}

1=(As+B)(s-1)² + C(s-1)(s²+1)+ D(s²+1)

The equation is reduced to:

1=s³(A+C)+s²(B-2A-C+D)+s(A-2B+C)+(B+D-C)

With the previous equation we can make an equation system of 4 variables.

The system is given by:

A+C=0

B-2A-C+D=0

A-2B+C=0

B+D-C=1

The solution of the system is:

A=1/2 ; B=0 ; C=-1/2 ; D=1/2

Therefore, Y(s) is equal to:

Y(s)=\frac{s}{2(s^{2} +1)} -\frac{1}{2(s-1)} +\frac{1}{2(s-1)^{2}}

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[\frac{s}{2(s^{2} +1)}]-ℒ⁻¹[\frac{1}{2(s-1)}]+ℒ⁻¹[\frac{1}{2(s-1)^{2}}]

y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}

8 0
3 years ago
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