Let x represent the number of shirts. Let y represent the number of pens.
If shirts are on sale for $11.99 each, then x shirts cost $11.99x.
If pants are on sale for $12.99 each, then y pants cost $12.99y.
The total cost is $(11.99x+12.99y).
Sarah can spend up to $65. Then an inequality that represents this situation is
11.99x+12.99y≤65 (this inequality holds when Sarah can spend $65 too)
or
11.99x+12.99y<65 (this inequality holds when Sarah can spend less than $65).
Firstly, since these two fractions have the same denominator, add the numerators up: 
While it appears that the answer is -6/4, you can further simplify it by dividing the numerator and denominator by 2: 
<u>Your answer is 
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Answer:
183.43x2
Step-by-step explanation:
Area of the entire frame = (14x)2 = 196x2
Area of the mirror = pi(2x)2 = 4x2(pi)
Area of the frame surrounding the mirror = (196 - 4pi)x2 = 183.43x2
Answer:
so basically
the area of the square; (5cm)^2 = 25cm^5
the area of the triangle; (4cm.5cm)/2 = 10cm^2
which means that r = √25cm^2/10cm^2 = √25/10 cm^2 or √10/25 cm^2
english is not my first language so i do not know what combined means but if u explain a little more i will try to help u
-> ok edit edit;
1cm/5in. = r
5cm => 25in.
z^2= area of a square so
25in.25in= 625 in^2
4cm => 20in.
5cm => 25in.
h.b/2 = area of a triangle so
20in.25in./2 = 250in.^2
250in.^2 + 625in.^2 = 875in.^2
so answer B
Answer:
Domain: {-6, -1, 7}
Range: {-9, 0, 9}
The relation is not a function.
Step-by-step explanation:
Given the relation: t{(−1,0),(7,0),(−1,9),(−6,−9)}
In the ordered pairs:
- The domain is the set of all "x" values
- The range is set of all "y" values
- We do not need to list any repeated value in the range/domain more than once.
Domain: {-6, -1, 7}
Range: {-9, 0, 9}
Next, we determine whether the relation is a function.
For a relation to be a function, each x must correspond with only one y value.
However, as is observed in the mapping attached below:
The x-value (-1) corresponds to two y-values (0 and 9)
Therefore, the relation is not a function.
