How do you do: -6(6a-5)=-2(8-5a)

Mathematics

1 answer:

devlian [24]4 years ago

7 0

Answer:

a = 1

Step-by-step explanation:

1. distribute the -6 on the left into the parentheses. you should get -36a + 30.

2. on the side right of the =, distribute the -2 to get -16 + 10a.

now it should look like -36a + 30 = -16 + 10a.

3. move the terms over to isolate the a value. subtract 30 to the right side to get -46 (-36a=-46+10a)

4. move the 10a on the right to the left by subtracting it to the other side.

(-46a=-46)

5.divide by -46 to get a=1

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Answer:

∠S = 66°

Step-by-step explanation:

A parallelogram's 4 angles always add up to 360°, and opposite angles are the same. (∠S = ∠U; ∠T = ∠V)

So, ∠S + ∠T = 180°.

180° = (2x + 4x + 12 + 6)°

180° = (6x + 18)°

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Assume that f is continuous on [-4,4] and differentiable on (-4,4). The table gives some values of f'(x) x: -4, -3, -2, -1, 0, 1

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$f$ will be increasing on the intervals where $f'(x)>0$ and decreasing wherever $f'(x)$ . Local extrema occur when $f'(x)=0$ and the sign of $f'(x)$ changes to either side of that point.

$f'(x)$ is positive when $x$ is between -4 and some number between -2 and -1, and also 2 (exclusive) and 4, so you can estimate that $f(x)$ is increasing on the intervals [-4, -2] and (2, 4].

$f'(x)$ is negative when $x$ is between some number between -2 and -1, up to some number less than 2. So $f(x)$ is decreasing on the interval [-1, 1].

You then have two possible cases for extrema occurring. The sign of $f'(x)$ changes for some $x$ between -2 and -1, and again to either side of $x=2$ .