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sladkih [1.3K]
3 years ago
14

Determine which of the lines, if any, are parallel. Explain.

Mathematics
1 answer:
algol133 years ago
7 0

Answer:

8) a and c

9) b and c

Step-by-step explanation:

Explanation and work are in the picture :)

You might be interested in
Solve these equation<br> 4(m+7)=44
Sphinxa [80]
4m+28=44
4m=44-28
4m=16
m=16/4
m=4
5 0
4 years ago
For example, if it takes 15 hours to travel 900 miles. How many miles will you travel in one hour?
dalvyx [7]

Answer:

60 miles

Step-by-step explanation:

If it takes 15 hours to travel 900 miles then it gives you the fraction 15/900 where the numerator(top number) is 15 and denominator(bottom number) is 900.

Then to simplify it: since 15 is a common factor of itself(15) and 900, 15/15 = 1(1 hour) and 900/15 = 60(60 miles).

8 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%
xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }

Let a = 1 and b the cosine product, and write them as

\dfrac{a - b}{x^2}

with

b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}

Now we use the identity

a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)

to rationalize the numerator. This gives

\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}

For the remaining limit, use the Taylor expansion for cos(x) :

\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)

where \mathcal{O}(x^4) essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2

\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)

\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0
3 years ago
The saddleback Seahawks football team gained 37 yards in the first quarter of the game, they lost 88 yards in the second Quarter
vivado [14]
Loss of yards

37 - 88 - 105 + 12 = -144
8 0
3 years ago
The scale on a map indicates that 3 inches equal to 9 miles. If the distance on the map between a city library and a school is 1
Nitella [24]

Answer: 45 miles.

Step-by-step explanation: 15 in/ 3in = 5 then take this 5 times the 9 miles = 45 miles.

5 0
4 years ago
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