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4 years ago

12

Find the limit of the function by using direct substitution. limit as x approaches one of quantity x squared plus three x minus

one.

1 answer:

ivann1987 [24]4 years ago

6 0

<u>Answer:</u>

Limit = -1

<u>Step-by-step explanation:</u>

We are given the following function:

$x^2 + 3x -1$

We are to calculate the limit of this function as x approaches zero.

For that, we will use direct substitution method and substitute the x with 0 in the given function to calculate its limit as follows:

$x ^ 2 + 3 x - 1$

$( 0 ) ^ 2 + 3 ( 0 ) - 1 = - 1$

Therefore, the limit is -1.

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Need some help! <br>help giving 10 points

steposvetlana [31]

Answer:

The remaining lenghts are $x = 20.364\,cm$ and $y = 12.925\,cm$ .

Step-by-step explanation:

Given that there are two triangles opposite to each other and between two parallel lines, then both are similar, meaning than following relationships exist:

$\frac{AC}{CD} = \frac{BC}{CE} = \frac{AB}{DE}$ (1)

If we know that $AC = 11\,cm$ , $CD = 28\,cm$ , $BC = 8\,cm$ and $DE = 32.9\,cm$ , then the missing lengths are, respectively:

$\frac{11\,cm}{28\,cm} = \frac{8\,cm}{x}$

$x = \frac{(28\,cm)\cdot (8\,cm)}{11\,cm}$

$x = 20.364\,cm$

$\frac{11\,cm}{28\,cm} = \frac{y}{32.9\,cm}$

$y = \frac{(11\,cm)\cdot (32.9\,cm)}{28\,cm}$

$y = 12.925\,cm$

The remaining lenghts are $x = 20.364\,cm$ and $y = 12.925\,cm$ .

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3 years ago

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3 years ago

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Neko [114]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Evaluate 4/3(4 1/5 + x) ÷ 5/2y when x = 3/2 and y = 15

MissTica

So what u do (-72828282)+(-7382828)=(-73838)+n=(-839288n)

8 0

3 years ago

I need help finding the inverse of this problem

fomenos

$y=\sqrt{x-1}+6$

(Notice that we always have $y\ge6[tex].)Swap [tex]x$ and $y$ , then solve for $y$ :

$x=\sqrt{y-1}+6$

$x-6=\sqrt{y-1}$

$(x-6)^2=(\sqrt{y-1})^2$

$(x-6)^2=y-1$

$y=(x-6)^2+1$ (this is the inverse)

###

This inverse is valid only for $x\ge6$ . Why?

Suppose we take $x=0$ . Then

$y=(0-6)^2+1=37$

This would suggest that in the original equation, we should get $x=37$ when $y=0$ . But when we check this, we end up with

$0=\sqrt{37-1}+6=\sqrt{36}+6=6+6=12$

which is clearly not true.

6 0

3 years ago