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TEA [102]
3 years ago
13

Consider the following equation. cos x = x3 (a) Prove that the equation has at least one real root. f(x) = cos x − x3 is continu

ous on the interval [0, 1], f(0) = 0, and f(1) = cos 1 − 1 ≈ −0.46 0. Since 0 −0.46, there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos x − x3 = , or cos x = x3, in the interval (0, 1). (b) Use your calculator to find an interval of length 0.01 that contains a root.

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
4 0

Answer:

b. 0.86, 0.87

Step-by-step explanation:

a. Find attached solution to a

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Answer:

49pi m^2

Step-by-step explanation:

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Please Help: Determine which ordered pair is NOT a solution of y = -5x - 4
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It's A.
All you need to do is plug in the X from the ordered pair, then solve.

y=-5x-4
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mars1129 [50]

Answer:

1.\:2^x=64, x=\boxed{6},\\2.\:x=\left(\frac{2}{3}\right)^3,x=\boxed{\frac{8}{125}}\\3.\:3\cdot (3^4)=3^x, x=\boxed{5}\\4.\:\frac{16}{25}=x^2,x=\boxed{\frac{4}{5}},

Step-by-step explanation:

1.\\\\2^x=64,\\\log 2^x=\log64,\\x\log 2=\log 64,\\x=\frac{\log 64}{\log 2}=\boxed{6}

2.\\x=\left(\frac{2}{5}\right)^3=\frac{2^3}{5^3}=\boxed{\frac{8}{125}}

3. This problem incorporates an exponent property.

Exponent property used: a^b\cdot a^c=a^{(b+c)}, yielding an answer of \boxed{5}

4.\\\\\frac{16}{25}=x^2,\\x=\sqrt{\frac{16}{25}}=\frac{\sqrt{16}}{\sqrt{25}}=\boxed{\frac{4}{5}}

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See attached files

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