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alexandr402 [8]
3 years ago
9

The ages of three cousins are consecutive even whole numbers. The sum of the youngest and oldest ages is 16. How old is the olde

st cousin?
Mathematics
2 answers:
mamaluj [8]3 years ago
3 0

the oldest cousin might be 20 years old

Liono4ka [1.6K]3 years ago
3 0

I believe the three ages are 6 8 & 10, the eldest being 10 and the youngest being 6.

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Simplify the expression.<br> 7x2 + 3 - 5(x2 - 4)<br> 2x2-17<br> 2x2-1<br> 2x2 + 23<br> 25x2
Blizzard [7]

Answer:

1. 14+3-5(-2)= 17+10=27

2. 14-17= -3

3. 4-1=3

4. 4+23=27

5. 50

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Fudgin [204]

Answer:

The ball reached its maximum height of (11\ yards) in (1\ second).

Step-by-step explanation:

This question is essentially asking one to find the vertex of the parabola formed by the given equation. One could plot the equation, but it would be far more efficient to complete the square. Completing the square of an equation is a process by which a person converts the equation of a parabola from standard form to vertex form.

The first step in completing the square is to group the quadratic and linear term:

h(t)=-5t^2 + 10t + 6\\\\h(t) = (-5t^2 + 10t) + 6

Now factor out the coefficient of the quadratic term:

h(t)=-5(t^2 -2t) + 6

After doing so, add a constant such that the terms inside the parenthesis form a perfect square, don't forget to balance the equation by adding the inverse of the added constant term:

h(t) = -5(t^2 -2t) + 6\\\\h(t) = -5(t^2 -2t + 1 -1 ) + 6

Now take the balancing term out of the parenthesis:

\\\\h(t)=-5(t^2 -2t + 1) + 6 + ((-1)(-5))

Simplify:

h(t) = -5(t^2 -2t + 1) + 6 + 5\\\\h(t) = -5(t-1)^2 + 11

The x-coordinate of the vertex of the parabola is equal to the additive inverse of the numerical part of the quadratic term. The y-coordinate of the vertex is the constant term outside of the parenthesis. Thus, the vertex of the parabola is:

(1, 11)

8 0
3 years ago
Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.
Aleks [24]

Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

Now:

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

if p>1 then series converges. In oue case we have:

\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

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Uninsured people are less likely to receive care and more likely to have poor health status.

Step-by-step explanation:

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3 years ago
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