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Nikolay [14]
3 years ago
15

What is (x^4)/(x^0)?

Mathematics
1 answer:
Lina20 [59]3 years ago
5 0

Answer:

\frac{x^{4}}{x^{0}}=\frac{x^{4}}{1}=x^{4}

Step-by-step explanation:

any~number~to~the~power~of~0~equal~to~1

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If the volume of the rectangular prism is represented by 6x^2-2x+8 and the base area is 2x-4 which expression represents the hei
MissTica
Check the picture below.

\bf 6x^2-2x+8=(2x-4)h\implies \cfrac{6x^2-2x+8}{2x-4}=h
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\cfrac{3(2x^2-x+4)}{2(x-2)}=h\implies \cfrac{3(3x-4)(x+1)}{2(x-2)}=h

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3 years ago
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If √6 is the geometric mean between 4 and another number, then the number is<br> 1.5<br> 9<br> 24
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Correct answer is A.

g - geometric mean
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4 0
3 years ago
Which angles are supplementary to each other?
Rus_ich [418]

Answer:

Angle 4 and Angle 1 are supplementary to each other.

Step-by-step explanation:

A line is 180°. Because angle 4 and angle 1 and right next to each other and they share a straight line, both of their angles should add up to 180°, making these two angles supplementary.

I hope this was helpful to you! If it was, please rate and press thanks! Have a fantastic day!

6 0
2 years ago
I NEED HELP ASAP PLEASE!
denis-greek [22]

Answer:

point X

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4 0
2 years ago
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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
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