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4 years ago

9

If a twelve hour clock is set at the correct time on the afternoon of may 17 if the clock loses 6 minutes per day what is the ne

xt calendar day on which it will show the correct time

1 answer:

Nookie1986 [14]4 years ago

3 0

Solution:

12 hours = 12 * 60 minutes = 720 minutes

Number of days lose = $\frac{720}{6} = 120 days$

So, 17th May + 120 days = 14th Sep.

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(D)Which equation matches the word problem? 24.35+45.65 = m

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given two terms in a geometric sequence find the 8th term and the recursive formula . a4=-12 and a5=-6

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A geometric sequence is defined by a starting point, $a$ , and a common ratio $r$

The first term is $a$ , and you get every next term by multiplying the previous one by r.

So, our terms are

$\left[\begin{array}{c|c}a_1&a\\a_2&ar\\a_3&ar^2\\a_4&ar^3=-12\\a_5&ar^4=-6\end{array}\right]$

We can see that when we pass from $a_4$ to $a_5$ the number gets halved ( $-12 \mapsto -6$ )

This implies that the common ratio is $r = \frac{1}{2}$

So, the table becomes

$\left[\begin{array}{c|c}a_1&a\\a_2&\frac{1}{2}a\\a_3&\frac{1}{4}a\\a_4&\frac{1}{8}a=-12\\a_5&\frac{1}{16}a=-6\end{array}\right]$

So, we can derive the starting point from either $a_4$ or $a_5$ :

$\dfrac{1}{8}a = -12 \iff a = -12\cdot 8 = -96$

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$\left[\begin{array}{c|c}a_1&-96\\a_2&-48\\a_3&-24\\a_4&-12\\a_5&-6\\a_6&-3\\\vdots&\vdots\end{array}\right]$

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4 years ago

Find, correct to the nearest degree, the three angles of the triangle with the given ven

Scrat [10]

9514 1404 393

Answer:

∠CAB = 86°

∠ABC = 43°

∠BCA = 51°

Step-by-step explanation:

This can be done a couple of different ways (as with most math problems). We can use the distance formula to find the side lengths, then the law of cosines to find the angles. Or, we could use the dot product. In the end, the math is about the same.

The lengths of the sides are given by the distance formula.

AB² = (4-1)² +(-3-0)² +(0-(-1)) = 16 +9 +1 = 26

BC² = (1-4)² +(2-(-3))³ +(3-0)² = 9 +25 +9 = 43

CA² = (1-1)² +(0-2)² +(-1-3)² = 4 +16 = 20

From the law of cosines, ...

∠A = arccos((AB² +CA² -BC²)/(2·AB·CA)) = arccos((26 +20 -43)/(2√(26·20)))

∠A = arccos(3/(4√130)) ≈ 86°

∠B = arccos((AB² +BC² -AC²)/(2·AB·BC)) = arccos((26 +43 -20)/(2√(26·43)))

∠B = arccos(49/(2√1118)) ≈ 43°

∠C = arccos((BC² +CA² -AB²)/(2·BC·CA)) = arccos((43 +20 -26)/(2√(43·20)))

∠C = arccos(37/(4√215)) ≈ 51°

The three angles are ...

∠CAB = 86°

∠ABC = 43°

∠BCA = 51°

_____

<em>Additional comment</em>

This sort of repetitive arithmetic is nicely done by a spreadsheet.

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