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3 years ago

What is this 8x+17=2x+35

1 answer:

7 0

Solve for x:
8 x + 17 = 2 x + 35
Subtract 2 x from both sides:
(8 x - 2 x) + 17 = (2 x - 2 x) + 35
8 x - 2 x = 6 x:
6 x + 17 = (2 x - 2 x) + 35
2 x - 2 x = 0:
6 x + 17 = 35
Subtract 17 from both sides:
6 x + (17 - 17) = 35 - 17
17 - 17 = 0:
6 x = 35 - 17
35 - 17 = 18:
6 x = 18
Divide both sides of 6 x = 18 by 6:
(6 x)/6 = 18/6
6/6 = 1:
x = 18/6
The gcd of 18 and 6 is 6, so 18/6 = (6×3)/(6×1) = 6/6×3 = 3:
Answer: x = 3

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A) Find a particular solution to y" + 2y = e^3 + x^3. b) Find the general solution.

Reptile [31]

Answer:

a.P.I= $\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)$

b.G.S= $C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x}$

Step-by-step explanation:

We are given that a linear differential equation

$y''+2y=e^{3x}+x^3$

We have to find the particular solution

P.I= $\frac{e^{3x}}{D^2+2}+\frac{x^3}{D^2+2}$

P.I= $\frac{e^{3x}}{3^2+2}+\frac{1}{2} x^3(1+\frac{D^2}{2})^{-2}$

P.I= $\frac{e^{3x}}{11}+\frac{1-2\frac{D^2}{4}+3\frac{D^4}{16}+...}{2}x^3$

P.I= $\frac{e^{3x}}{11}+\frac{x^3-2\frac{\cdot3\cdot 2x}{4}}{2}+0}$ (higher order terms can be neglected

P.I= $\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)$

b.Characteristics equation

$D^2+2=0$

$D=\pm\sqrt2 i$

C.F= $C_1cos \sqrt2x+C_2 sin\sqrt2 x$

G.S=C.F+P.I

G.S= $C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x)$

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3 years ago

4 friends went to the movies and each ordered a container of popcorn.They spent $24 on popcorn.How much did each container of po

alexandr402 [8]

Answer: $6

Step-by-step explanation:

24/4 = 6

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3 years ago

Read 2 more answers

What is the difference in solving functions when giving the following information: f(x)=2 and f(2)

vfiekz [6]

<em>*Remember that y and f(x) are the same.</em>

With f(2), you are solving for the output (y), given that the input (x) is equal to 2.

With f(x) = 2, you are solving for the input (x), given that the output (y) is equal to 2.

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3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

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3 years ago

the ratio of boys to girls at the beach clean up was 7:8 if there are 42 boys how many girls are there

Len [333]

48 girls are there. Given that there is a ratio of boys to girls of 7:8. Then divide 42 boys by 7 boys which equals 6. After, multiply 6 by 8 girls equals 48 girls.

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