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Lerok [7]
3 years ago
8

I need help on #2 plz help

Mathematics
1 answer:
Ainat [17]3 years ago
5 0

Answer:

Please see attached picture for full solution.

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Force is giving away tickets to a school play. She gives away 20 more adult tickets than children's tickets. She gives away 3 ti
swat32
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8 0
3 years ago
If a runner was traveling with a velocity of 4 m/s north and had a displacement of 2500 m, how long was he running?
Mademuasel [1]

Answer:

625

Step-by-step explanation:

t= d/s

2500/4

4 0
3 years ago
Which statements about the function are true? Select two
iogann1982 [59]

Answer:

The vertex of the function is at (1,-25)

Step-by-step explanation:

I think your question missed key information, allow me to add in and hope it will fit the orginal one.

<em>Part of the graph of the function f(x) = (x + 4)(x-6) is shown  below. </em>

<em>Which statements about the function are true? Select two </em>

<em>options. </em>

<em>The vertex of the function is at (1,-25). </em>

<em>The vertex of the function is at (1,-24). </em>

<em>The graph is increasing only on the interval -4< x < 6. </em>

<em>The graph is positive only on one interval, where x <-4. </em>

<em>The graph is negative on the entire interval  </em>

My answer:

Given the factored form of the function:

f(x) = (x + 4)(x-6)

<=> f(x) = x^{2} - 2x -24

We will convert to vertex form

<=> f(x) = (x^{2} - 2x +1) - 25

<=> f(x) = (x-1)^{2} -25

=> the vertex of the function is: (1,-25)

We choose: a. The vertex of the function is at (1,-25)

Let analyse other possible answers:

<u>c. The graph is increasing only on the interval -4< x < 6.</u>

Because the parameter a =1 so the graph open up all over its domain and the vertex is the lowest point.

So the graph is increasing in the domain (1, +∞)

=> C is wrong

<u>d. The graph is positive only on one interval, where x <-4</u>

Wrong, The graph is positive only on one interval, where x > 6

<u>e. The graph is negative on the entire interval</u>

Wrong, The graph is negative only on one interval, where -4< x < 6.

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4 years ago
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Whats the starting point
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4 years ago
What is lim x → 0 e^2x - 1/ e^x - 1
olya-2409 [2.1K]

Hello, please consider the following.

\displaystyle \forall x \in \mathbb{R}\\\\\dfrac{e^{2x}-1}{e^x-1}\\\\=\dfrac{(e^x)^2-1^2}{e^x-1}\\\\=\dfrac{(e^x-1)(e^x+1)}{e^x-1}\\\\=e^x+1\\\\\text{So, we can find the limit.}\\\\\lim_{x\rightarrow 0} \ {\dfrac{e^{2x}-1}{e^x-1}}\\\\=\lim_{x\rightarrow 0} \ {e^x+1}\\\\=e^0+1\\\\\large \boxed{\sf \bf \ =2 \ }

Thank you

4 0
4 years ago
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