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Genrish500 [490]
3 years ago
14

8x - 3y, using x = 5 and y = 2

Mathematics
1 answer:
jeka57 [31]3 years ago
5 0

Step 1: plug in 5 and 2 for your X and Y

Step 2: multiply 8*X or in this case 8*5 and you will get 40

Step 3: multiply 3*Y or in this case 3*2 and you will get 6

Step 4: subtract 40-6 and you will get 34

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6^2 + x - x^2 for x=3 <br> help me pls
Stels [109]

Answer:

30

Step-by-step explanation:

Just insert the value of x into the equation.

x = 3 right?

6^2 + 3 - 3^2 because x = 3

36 + 3 - 9 = 39 -9 = 30

:D

4 0
3 years ago
Plz help me I’m not good at math
grigory [225]

Answer:

second one

Step-by-step explanation:

am pretty sure

6 0
3 years ago
Read 2 more answers
Permutations: 5!·4! is equivalent to what (5!)(4!)^2, (5)(4!)^2 or 20! ?
solniwko [45]

Answer:

5!*4! is equivalent to 5*(4!)^2

Step-by-step explanation:

5!*4! = 2880

5*(4!)^2 = 2880

4 0
3 years ago
Which expression is equivalent to -14 - 35d
Lostsunrise [7]
The first one I believe
8 0
3 years ago
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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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