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3 years ago

Solve the system by elimination 2x+6y= -12 5x-5y= -10

2 answers:

5 0

First thing you should do is reduce coefficients.

1st equation has all multiples of '2'. Divide by 2
---> x +3y = -6

2nd equation has multiples of 5. Divide by 5.
---> x - y = 2

Now elimination part is easier.
Eliminate 'x' variable by subtracting 2nd equation from 1st.

x + 3y = -6
-(x - y = 2)
----------------------
4y = -8

Solve for 'y'
4y = -8

y = (-8)/4 = -2

Substitute value for 'y' back into 2nd equation:
x - (-2) = 2
x + 2 = 2
x = 0

Solution to system is:
x=0, y =-2

Lena [83]3 years ago

3 0

B) (0,-2) I got 100% right.

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Find all values of c in the open interval (a, b) such that f'(c)=(f(b)-f(a))/(b-a)

timama [110]

<h3>Answer: c = 7/4</h3>

================================================

Work Shown:

Compute the function value at the endpoints

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With a = -5 and b = 4, we have

$f'(c) = \frac{f(b)-f(a)}{b-a}\\\\f'(c) = \frac{f(4)-f(-5)}{4-(-5)}\\\\f'(c) = \frac{0-3}{9}\\\\f'(c) = -\frac{1}{3}\\\\$

So,

$f(x) = \sqrt{4-x}\\\\f'(x) = -\frac{1}{2\sqrt{4-x}}\\\\f'(c) = -\frac{1}{3}\\\\-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\$

Use algebra to solve for c

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mars1129 [50]

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Step-by-step explanation:

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Ne4ueva [31]

Answer:

all non-negative real numbers, [0, ∞)

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