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3 years ago

What is the greatest value in the range of f(x)=x^2-5 for the domain {-5, -3, 0, 1, 2}?

1 answer:

3 0

$f(x)=x^2-5;\ x\in\{-5;-3;\ 0;\ 1;\ 2\}\\\\f(-5)=(-5)^2-5=25-5=20\\f(-3)=(-3)^2-5=9-5=4\\f(0)=0^2-5=-5\\f(1)=1^2-5=1-5=-4\\f(2)=2^2-5=4-5=-1\\\\Answer:f(-5)=20$

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Graph the equation y= x^2 - 2x - 8 on the accompanying set of axes. You must plot 5 points including the roots and the vertex.

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Answer:

I'm not so sure about this one so sorry if this is wrong

Step-by-step explanation:

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3 years ago

Solve equation by completing the square 5n^2-3n-10=15

VARVARA [1.3K]

These are the two answers you could get

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Function f(x) is positive, increasing and concave up on the closed interval [a, b]. The interval [a, b] is partitioned into 4 eq

Umnica [9.8K]

The left sum would be f0+f1+f2+f3

The right sum would be f1+f2+f3+f4

The trapezoidal rule value is:

(f0+f1)/2 + (f1+f2)/2+(f2+f3)/2 +(f3+f4)/2

This would put the trapezoidal rule in the middle , which makes the answer:

Lower sum < Trapezoidal rule Value < Upper sum

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3 years ago

Read 2 more answers

If a particular set of data is approximately normally distributed, we would find that approximately A) 2 of every 3 observations

Kryger [21]

Answer:

A) No

B) No

C) Yes

Step-by-step explanation:

A) 2 out of three observations is equal to

2/3 = 0.67 or 67 %

We know tha the interval

[μ - σ/2 , μ + σ/2] is approximately 68.3 % of the values

Therefore 2 out of 3 would not fall in that interval

4 out of 5 is 4/5 = 0,8 or 80 % If we examine the interval

[μ - σ , μ + σ ] which is approximately 95. Then value 4/5 ( 0,8)

Therefore 0,8 is out of the interval

C) 19/20 = 0,95 or 95 % is almost exact the value for the interval

[μ - 1.5σ , μ + 1.5σ ] That value is inside μ - 2σ , μ + 2

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3 years ago

Elise walks diagonally from one corner of a square plaza to another. Each side of the plaza is 50 meters. What is the diagonal d

jolli1 [7]

Answer:

70.7 meters.

Step-by-step explanation:

We have been given that Elise walks diagonally from one corner of a square plaza to another. Each side of the plaza is 50 meters.

Since we know that diagonal of a square is product of side length of square and $\sqrt{2}$ . So we will find diagonal of our given square plaza by multiplying 50 by $\sqrt{2}$ .

$\text{Diagonal distance across the plaza}=50\times \sqrt{2}$

$\text{Diagonal distance across the plaza}=50\times 1.414213562373095$

$\text{Diagonal distance across the plaza}=70.71067811865475\approx 70.7$

Therefore, diagonal distance across the plaza is 70.7 meters.

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3 years ago