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Anestetic [448]
3 years ago
13

Write the equation of a line that includes the point (1, 5) and has a slope of 3 in standard form.

Mathematics
1 answer:
ale4655 [162]3 years ago
4 0
Y = mx + b
m = 3 (given), then

y = 3x + b. Now calculate b by plugging in the coordinates (1,5)

5 = 3(1) + b and b = 2

The equation is then y = 3x+2

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natali 33 [55]
3/5 + 5/7 = 1 11/35 - 1/5 = 1 4/35 which is your final answer.
7 0
3 years ago
Read 2 more answers
X+5=5<br> What is the value?
AveGali [126]
Answer: 1x or x

X=5-5
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3 years ago
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What is the answer for 2/5x5/8 ?
Maru [420]
It would be 1/4 because you would change them to a common denominator then simply them and you get 1/4
7 0
3 years ago
What is the value of x in the equation below?<br><br> 6 (x minus 8) = 72
9966 [12]

Answer:

x = 20

Step-by-step explanation:

6(x - 8) = 72 \\ 6x - 48 = 72 \\ 6x = 120 \\ x = 20

5 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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