3 years ago

Solve for a. Show your work plz.

Mathematics

1 answer:

NeTakaya3 years ago

6 0

Answer:

Step-by-step explanation:

$\frac{10}{a} =sin 60\\a=\frac{10}{sin~60}\\=\frac{10}{\frac{\sqrt{3} }{2} } \\=\frac{20}{\sqrt{3}} \times\frac{\sqrt{3}}{\sqrt{3}} \\=\frac{20 \sqrt{3}}{3}$

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Find slope through (3,16) and (8,18)

MissTica

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3 years ago

4 lines are shown. 3 lines intersect and extend from one point forming 2 angles. The top angle is labeled 1 and the bottom angle

svp [43]

Answer:

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Step-by-step explanation:

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3 years ago

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Let f be a continuous function defined on the interval [0,1] such that f(0) = f(1). Show that there exists a number 0 <= a &l

Leni [432]

Explanation:

In order to prove that affirmation, we define the function g over the interval [0, 1/2] with the formula $g(x) = f(x+1/2)-f(x) .$

If we evaluate g at the endpoints we have

g(0) = f(1/2)-f(0) = f(1/2) - f(1) (because f(0) = f(1))

g(1/2) = f(1) - f(1/2) = -g(0)

Since g(1/2) = -g(0), we have one chance out of three

g(0) > 0 and g(1/2) < 0
g(0) < 0 and g(1/2) > 0
g(0) = g(1/2) = 0

We will prove that g has a zero on [0,1/2]. If g(0) = 0, then it is trivial. If g(0) ≠ 0, then we are in one of the first two cases, and therefore g(0) * g(1/2) < 0. Since f is continuous, so is g. Bolzano's Theorem assures that there exists c in (0,1/2) such that g(c) = 0. This proves that g has at least one zero on [0,1/2].

Let c be a 0 of g, then we have

$0 = g(c) = f(c+1/2)-f(c)$