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3 years ago

I need to find the domain and ranges of some of these graphs and I can’t really do the last few! Can anyone help?

Mathematics

1 answer:

hodyreva [135]3 years ago

3 0

(1) For the parabola on the bottom row, the domain would be R and the range would be y ≥ -5

(2) For the hyperbola on the bottom row, the domain would be R\{3} (since there is an asymptote at x = 3) and the range would be R\{4} (since there is an asymptote at y = 4)

(3) For the square root function on the bottom row, the domain would be x ≥ -5 and the range would be (-∞, -2]

(4) For the function to the very right on the bottom row, the domain would be R and the range would be (-∞, -3]

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Answer:13 to the 5th power

Step-by-step explanation:

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3 years ago

Read 2 more answers

The perimeter of the triangular park shown on the right is 10x - 3. What is the missing length?

viva [34]

Answer:

6x - 1

Step-by-step explanation:

Okay this is basically polynomial algebra:

1. Add the value of the lengths already given.

like terms: 3x + x = 4x, -4 + 2 = -2. Now you're left with, 4x - 2.

2. Find the difference between the perimeter and the sum of the lengths.

(10x -3) + (4x + 2) = 6x - 1

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3 years ago

The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas

Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is $= \frac{(n+1)}{2}$ .

$\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5$ position orders

$10^{th} \ and \ 11^{th}$ place an average of observations so, the

$Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430$

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

$Q_1$ often falls throughout the ordered BELOW average is $= \frac{(n+1)}{2}$ place.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered position

Average $5^{th} \ and \ 6^{th}$ location findings

$Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5$

In $= \frac{(n+1)}{2}$ the ordered place, Q3 always falls ABOVE the median.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered At median ABOVE

Consequently $Q_3$ falls between $5^{th} \ and \ 6^{th}$ ABOVE the median position

$15^{th}\ and \ 16^{th}$ place average of observations

$\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2} = 502$

$\to IQR = Q_3 - Q_1= 502-358.5= 143.5$

$\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25$

4 0

3 years ago

HELP PLEASE MARK BRAINLIEST ANSWER PLEASE HELP ME THE GRAPH IS IN THE PICTURE THANK YOU AND PLEASE REALLY HELP ME SOMEONE THAT I

kipiarov [429]

Answer:

The answer would be G.

Step-by-step explanation.

To get your answer, you have to pick two points. Ex: (0,4) (4,1)

Put them into the slop equation: slope= y2-y1/x2-x1

slope:1-4/4-0= -3/4

5 0

3 years ago