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3 years ago

Evaluate the expression when x=-6. x² + 5x-5

Mathematics

1 answer:

Rus_ich [418]3 years ago

6 0

Answer:

Step-by-step explanation:

x² + 5x-5

Let x = -6

(-6)^2 + 5(-6) -5

36 -30 -5

6-5

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4(1 - x) + 3x = - 2(x + 1)

liubo4ka [24]

Answer:

X=-6

Step-by-step explanation:

4(1−x)+3x=−2(x+1)

(4)(1)+(4)(−x)+3x=(−2)(x)+(−2)(1)(Distribute)

4+−4x+3x=−2x+−2

(−4x+3x)+(4)=−2x−2(Combine Like Terms)

−x+4=−2x−2

Step 2: Add 2x to both sides.

−x+4+2x=−2x−2+2x

x+4=−2

Step 3: Subtract 4 from both sides.

x+4−4=−2−4

x=−6

Answer:

x=−6

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In geometric progression, the ratio of each number to the preceding one is the same. Which is an example of a geometric progress

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It would be A because this sequence has a factor of 2 between each number

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solmaris [256]

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2 years ago

The expression cos^-1 (3/5) has an infinite number of values.. True or False.

Pavel [41]

Answer:

The given statement:

The expression cos^-1 (3/5) has an infinite number of values is a true statement.

Step-by-step explanation:

We are given a expression as:

$\arccos (\dfrac{3}{5})$

Let us equate this expression to be equal to some angle theta(θ)

i.e.

Let

$\arccos (\dfrac{3}{5})=\theta\\\\\cos \theta=\dfrac{3}{5}$

As we know that the limit point of the cosine function is [-1,1]

i.e. it takes the value between -1 to 1 and including them infinite number of times.

Also,

-1< 3/5 <1

This means that the cosine function takes this value infinite number of times.

That is there exist a infinite number of theta(θ) for which:

$\cos \theta=\dfrac{3}{5}$

i.e. the expression:

$\arccos (\dfrac{3}{5})$ has infinite number of values.

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