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3 years ago

I need to evaluate h(3)

Mathematics

1 answer:

Natali5045456 [20]3 years ago

8 0

Answer:

h(3) = - 1

Step-by-step explanation:

To evaluate h(3), that is x = 3 in the interval x ≥ 1 , then

h(x) = - x + 2 , thus

h(3) = - 3 + 2 = - 1

Then h(3) = - 1

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A = (12*3) + (4*9) = 72 cm^2

Step-by-step explanation:

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2 years ago

Read 2 more answers

Hurry 50 points for correct answer

Ad libitum [116K]

Answer: <em>"7 is a solution to the original equation. The value –1 is an extraneous solution."</em>

Step-by-step explanation:

The equation $x=\sqrt{6x+7}$ can be solved by squaring both sides:

$x^2 = 6x + 7\\\\x^2 - 6x - 7 = 0\\\\(x-7)(x+1) = 0$

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$1 = \sqrt{-1}\\ 7 = \sqrt{49}$

The square root of 49 equals 7, but the square root of -1 is an imaginary number.

The correct choice is <em>"7 is a solution to the original equation. The value –1 is an extraneous solution."</em>

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3 years ago

If my rounded number is 9.7 what could the original number have been

castortr0y [4]

Answer:

9.65-9.74

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2 years ago

Very confused especially on number three.

kogti [31]

Answer:

yes it was so confused for me too

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3 years ago

What equation is graphed in this figure?

topjm [15]

Answer:

The third equation: $y+2=-3(x-1)$

Step-by-step explanation:

The two points on the line are $(-1,4)$ and $(1,-2)$ .

Slope of the line passing through two points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given as:

$m=(y_{2} -y_{1})/ (x_{2} -x_{1})$

Here, $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are $(-1,4)$ and $(1,-2)$ .

Therefore, slope is equal to, $m=(-2-4 )/ (1-(-1)$

$m=-6/2$

$m=-3$

Now, equation of a straight line with slope m and points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given as:

$y-y_{1}=m(x-x_{1})\\y-y_{2}=m(x-x_{2})$

Now, if we use the 2nd form, then $x_{2}=1,y_{2}=-2$ .

So, the equation is given as :

$y-(-2)=-3(x-1)\\y+2=-3(x-1)$

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3 years ago