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kykrilka [37]
3 years ago
8

If y varies inversely as x, and y = 4 as x = 8, find y for the x-value of 2.

Mathematics
2 answers:
alexira [117]3 years ago
8 0

y = k/x

We want to find k first.

4 = k/8

4(8) = (k/8)(8)

32 = k

We now need to find y knowing that k = 32 and x = 2.

y = k/x

y = 32/2

y = 16

Did you follow?

cricket20 [7]3 years ago
6 0

Answer:

16

Step-by-step explanation:

Inverse variation follows this equation:

y = k/x

We know one point, so we input x and y, and solve for k.

4 = k/8

k = 32

The inverse variation equation is

y = 32/x

Now we let x = 2, and use our equation to find y.

y = 32/2

y = 16

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if A and B are overlapping subset of a universal set U,write the relation between n(U), n(A U B) and (A U B) compliments​
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Answer:

n(U) = n(AUB) + n(AUB)'

Step-by-step explanation:

The cardinality of the universal set will be to add the cardinalities of both the union of A and B i.e (A U B) and and the compliments of A and B i.e n(AUB)'

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3 years ago
What is the solution?
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irina1246 [14]

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Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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)

3 0
3 years ago
Thirty-two cyclists make a seven-day trip. Each cyclist requires 8.33 kilograms of food for the entire trip. If each cyclist wan
kvasek [131]

Answer:

76.16 kilograms of food

Step-by-step explanation:

Number of cyclist = 32

Food per cyclist = 8.33 kilograms

Days of the trip = 7 days

Food per cyclist per day = 8.33 kg / 7

= 1.19kg food per day for each cyclist

how many kilograms of food will the group be carrying at the end of Day 5?

5 days out of 7 days = 2 days

Each cyclist will have to carry = 2 × 1.19 kg of food for the remaining two days

= 2 × 1.19

= 2.38 kilograms of food for two days

32 cyclist for 2 days = 32 × 2.38 kg of food

= 76.16 kilograms of food

The group will be carrying 76.16 kilograms of food at the end of Day 5

3 0
3 years ago
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