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Pachacha [2.7K]
3 years ago
12

18x-6y= 54 12y=6x-18 Solve this system of equations using the substitution method.

Mathematics
1 answer:
olga_2 [115]3 years ago
5 0

Answer:

x=10/3 , y=1/6

Step-by-step explanation:

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Determine if the function f(x)=44−x2‾‾‾‾‾‾√ satisfies the Mean Value Theorem on [0, 2]. If so, find all numbers c on the interva
lina2011 [118]

Answer:

The function f(x) satisfies the Mean Value Theorem

Step-by-step explanation:

Mean Value Theorem states that if f be a function such that,

  •  It is continuous on [a, b].
  • It is differentiable on (a, b).

Then there is at least a number c in (a, b) such that,

f'(c) = \frac{f(b) - f(a)}{b-a}

Here, the given function,

f(x) = \sqrt{44-x^2}

∵ f(x) is defined for all real values x for which 44-x² ≥ 0

44 ≥ x² ⇒ ±√44 ≥ x ⇒-√44 ≤ x ≤ √44

Thus, f(x) is continuous on [0, 2],

f'(x) = \frac{1}{2\sqrt{44-x^2}}(-2x) =-\frac{x}{\sqrt{44-x^2}}

∵ f'(x) is defined for all values on the interval (0, 2),

Thus, f(x) is differentiable on (0, 2),

Now,

f'(c) = -\frac{c}{\sqrt{44-c^2}}

\frac{f(2) - f(0)}{2-0}=\frac{\sqrt{44-4}-\sqrt{44}}{2}=\frac{\sqrt{40}-\sqrt{44}}{2}=\sqrt{10}-\sqrt{11}

-\frac{c}{\sqrt{44-c^2}}=\sqrt{10}-\sqrt{11}

\frac{c^2}{44-c^2}=10 + 11 - 2\sqrt{110}

c^2 = (44-c^2)(21-2\sqrt{110})

c^2 = 924 - 88\sqrt{110} - c^2(21 - 2\sqrt{110})

c^2 + (21 - 2\sqrt{110})c^2 = 924 - 88\sqrt{110}

(22 - 2\sqrt{110})c^2 = 924 - 88\sqrt{110}

c^2 = \frac{924 - 88\sqrt{110}}{22 - 2\sqrt{110}}

c=\sqrt{ \frac{924 - 88\sqrt{110}}{22 - 2\sqrt{110}}}\approx 1.012\in (0, 2)

Hence, the function f(x) satisfies the Mean Value Theorem

7 0
4 years ago
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