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Leni [432]
2 years ago
12

Find the distance of the line segment joining the two points (5, 4) and (−2, 1)

Mathematics
1 answer:
Scilla [17]2 years ago
3 0
Distance formula... but the answer is square root of 58 units with is approximately 7.61577
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HELPP
LenaWriter [7]
F, D, C
F: 5 x 3 = 15
D: 7 x 3= 21 - 2 = 19
C: 5 x 3 = 15 + 6 = 21
6 0
2 years ago
Read 2 more answers
HELPPP?!!! What is the value of f(-2) and f(3)
Cerrena [4.2K]
Step #1 for both: figure out which interval your x-value fits into.
For f(-2), x=-2 and -2 fits with x ≤ -2, the top interval.
For f(3), x=3 and 3 fits into -2 < x ≤ 3, the middle interval.

Step #2 for both, plug in your x-value to the piece of the function that fits with that interval.
For f(-2), we know x≤-2, so we use 2x+8 to evaluate x=-2.
For f(3), we know -2
f(-2) = 2(-2)+8 = -4+8 = 4

f(3) = (3)^2 -3 = 9-3 = 6
3 0
2 years ago
Andrew borrowed $700 from the bank to start a pressure washing business. The bank charged a simple interest rate of 7% per year.
Flauer [41]

Answer:6%

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
What’s 9 times 9<br><br> Just gave you 100 for free your welcome
Rudiy27

Answer:

81!    THANK YOU<3333333333333

6 0
2 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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