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3 years ago

If the relationship between x and y is quadratic,

Mathematics

1 answer:

maxonik [38]3 years ago

4 0

I am not sure but I think I know ok

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At a picnic there were 3 times as many adults as children and twice as many women as men. If there was a total of x men, women,

Dennis_Churaev [7]

Answer:

$\dfrac{x}{4}$

C is correct.

Step-by-step explanation:

At a picnic,

Number of adults is 3 times as number of children.

Number of women is twice as number of men.

Total number of men, women and children at the picnic be x

Let number of children be c

Let number of men be m

Let number of women be w

# Number of women is twice as number of men, w = 2m

# Number of adults is 3 times as number of children, w + m = 3c

2m + m = 3c (∴ w=2m )

c = m

Total number of men, women and children at the picnic be x

∵ c + m + w = x

m + m + 2m = x

4m = x

Number of men, $m=\dfrac{x}{4}$

Hence, The total number of men will be $\dfrac{x}{4}$

5 0

2 years ago

Why can you NOT combine 3x and 5x^2 ?

diamong [38]

Although they are the same variable, you cannot add two variables raised to different powers. 3x^1 + 5x^2 cannot work, but 3x^2 + 5x^2 can. Also, if you are multiplying them, they would combine to be 15x^3, as 3 and 5 (The coefficients) multiply together and x^2 times x^1 = x^3.

I hope that helps.

7 0

3 years ago

HELP I WILL GIVE YOU 20 POINTS

shepuryov [24]

Maby c I may be wrong

5 0

3 years ago

Read 2 more answers

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago

Find the first 5 terms of the arithmetic sequence : a(1)=-3 , a(17)=-20

madreJ [45]

The explicit formula for arithmetic sequence is:
an=a+(n-1)d
where:
a=first term
d=common difference
given:
a3=22
a(17)=-20
substituting this in our equation we get:
22=a+(3-1)d
22=a+2d
a=22-2d........i

also:
-20=a+(17-1)d
-20=a+16d.....ii
but substituting i in ii we get:
-20=22-2d+16d
-20-22=14d
-42=14d
d=-3
but:
a=22-2d
a=22-2(-3)
a=28
thus the formula will be:
an=28-3(n-1)
thus the first term will be 28
the 2nd term will be:
a2=28-3(2-1)
a2=25

the 3rd term will be:
a3=28-3(3-1)
a3=28-6
a3=22

a4=28-3(4-1)
a4=28-9
a4=15

a5=28-3(5-1)
a5=28-3(4)
a5=28-12
a5=15

5 0

2 years ago