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OLEGan [10]
3 years ago
8

Which function has a domain where x=3 and a range where y=2 ?

Mathematics
2 answers:
BabaBlast [244]3 years ago
8 0

Answer:

x times y is 6

This is the simplest one

Step-by-step explanation:

x=3 and y=2

Elza [17]3 years ago
4 0

Answer:

(3,4)

Step-by-step explanation:

so domain is the x value and range is the y value so to find the domain it is

(3,4)

Hope that helps :)

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The circumference of a circle is 26 pi what is the area in square inches of the circle
melisa1 [442]

circumference=

2\pi \: radius

26\pi = 2\pi \: raduis \:

\frac{26\pi}{2\pi}  = raduis

radius = 13

area of circle =

\pi {raduis}^{2}

area of circle=

\pi {13}^{2}

area of circle=

169\pi

8 0
3 years ago
The slope of the line whose equation is 3y + 2x = 1 is: A.)-2/3 B.)-3/2 C.)1/3
mojhsa [17]

Answer:

A

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Rearrange 3y + 2x = 1 into this form

Subtract 2x from both sides

3y = - 2x + 1 ( divide all terms by 3 )

y = - \frac{2}{3} x + \frac{1}{3} ← in slope- intercept form

with slope m = - \frac{2}{3} → A

4 0
3 years ago
Evaluate.
kogti [31]
9/10 - 2/5 + 1/3

= 5/6 (Decimal:  0.8333333)


Subtract:
9/10 - 2/5 = 9/10 - 2 . 2/5 . 2
= 9/10 - 4/10 
= 9 - 4/10 
= 5/10 
= 1/2

The common denominator you can calculate as the least common multiple of the both denominators LCM ( 10, 5) = 10


Add: 1/2 + 1/3 = 1 . 3/2 . 3 + 1 . 2/3 . 2 
= 3/6 + 2/6 
= 3 + 2/6 

= 5/6



The common denominator you can calculate as the least common multiple of the both denominators LCM ( 2, 3) = 6

Answer in Lowest term is:  5/6 or Decimal:  0.8333333




Hope that helps!!!




7 0
3 years ago
Line l and line m intersect.<br>Prove: ∡1 ≅ ∡3
ikadub [295]
Line l and line m intersect
2 is supplementary to 3
Angle 1 congruent to angle 3

Hope this helped :)
5 0
2 years ago
Read 2 more answers
Find the length of base ab of trapezoid abcd
vesna_86 [32]
By doing base times area that is what I will do
6 0
4 years ago
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