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3 years ago

(For both) I will give brainliest

Mathematics

2 answers:

aev [14]3 years ago

4 0

Answer:

1. yes

2. no

Step-by-step explanation:

amm18123 years ago

3 0

The first is yes and the second is no

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The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula below:

Anni [7]

Given $y_1=e^{2x}$ is a fundamental solution, we posit a second solution of the form $y_2=y_1v=e^{2x}v$ , with derivatives

${y_2}'=e^{2x}v'+2e^{2x}v=e^{2x}(v'+2v)$

${y_2}''=e^{2x}v''+4e^{2x}v'+4e^{2x}v=e^{2x}(v''+4v'+4v)$

Substitute these into the ODE:

$e^{2x}(v''+4v'+4v)-4e^{2x}(v'+2v)+4e^{2x}v=0\implies v''=0$

Integrate both sides twice to get

$v''=0\implies v'=C_1\implies v=C_1x+C_2$

Then the second fundamental solution is

$y_2=xe^{2x}+e^{2x}$

but $y_1$ already cover $e^{2x}$ , so $y_2=xe^{2x}$ .

6 0

3 years ago

Benji took out a $10,000 loan at a 13% interest rate. Benji will not make a payment for the next 3 years. How much will be owed

xxMikexx [17]

Answer:

Plug into calc

Step-by-step explanation:

A= P(1+r/n)^nt

A= 10000(1+.13/12)^12*3

7 0

3 years ago

The sum of two numbers is -1. When twice the first number and four times the second number are added, it equals -10. What are th

wlad13 [49]

Let the two number be x and y.
x+y=-1......i
2x+4y=-10.....ii
solving the simultaneous equation we have:
from equation i, making x the subject we get:
x=-1-y........iii
substituting iii in ii we get:
2(-1-y)+4y=-10
-2-2y+4y=-10
2y=-10+2
2y=-8
y=-8/2=-4
substituting the value of y in iii we get:
x=-1-(-4)
x=-1+4
x=3
therefore the first number is 3 and the second number is -4

3 0

3 years ago

A cube has side length a. The side lengths are decreased to 50% of their original size. Write an expression in simplest form for

vaieri [72.5K]

Answer:

V = a³/8

Step-by-step explanation:

The volume of the original cube is the cube of the side length:

V = a³

When the side length is reduced to half its former value, the new volume is ...

V = (a/2)³ = a³/2³

V = a³/8 . . . . volume of the new cube