Answer:
Which letter has two lines of symmetry? Y S M X plzz help due tomorrow
2 answers:
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Answer:
34) A
35) C
36) D
Step-by-step explanation:
34) The perimeter of a triangle is equal to the sum of the lengths of its sides
Sides: (6.2p + 4.5q) + (3.4p - 0.7r) + (2.9r - 4.1q)
Combine like terms (same variables)
(6.2p + 3.4p) + (4.5q - 4.1q) + (2.9r - 0.7r)
9.6p + 0.4q + 2.2r
A
35)
Time possibilities
1 hour - $3 - $5 = $-2 No
2 hours - $6 - $5 = $1 Yes
3 hours - $9 - $5 = $4 Yes
4 hours - $12 - $5 = $7 Yes
5 hours - $15 - $5 = $10 Yes
6 hours - $18 - $5 = $13 Yes
7 hours - $21 - $5 = $16 Yes
8 hours - $24 - $5 = $19 No
9 hours - $27 - $5 = $22 No
The possible hours that Ms. Barnes can park are 2 - 7 hours
The only inequality graph that represents all of these values is C
36)
<h3>Convert to improper fractions
<h3>Divide by finding the reciprocal
<h3>Convert back to a mixed number
D
Hope this helps :)
The equation is 
.
We are looking for a function with a vertex above the x-axis and a function that opens upward (has coefficient a > 0).
The first function opens downward and intersects the x-axis. The second function has a vertex below the x-axis. The third function satisfies our requirements. The fourth function has a vertex on the x-axis.
We can solve this algebraically with the knowledge that the real solutions of a quadratic are its x-intercepts. If there are no x-intercepts (because it lies entirely above or below the x-axis), then there are no real solutions. This is true when the discriminant
. You can see that from the quadratic formula. This holds true for both answers A and C, so to find the correct one, we remember that when the coefficient a of the 
term is positive, the graph opens upwards, so we choose C.
We are looking for a function with a vertex above the x-axis and a function that opens upward (has coefficient a > 0).
The first function opens downward and intersects the x-axis. The second function has a vertex below the x-axis. The third function satisfies our requirements. The fourth function has a vertex on the x-axis.
We can solve this algebraically with the knowledge that the real solutions of a quadratic are its x-intercepts. If there are no x-intercepts (because it lies entirely above or below the x-axis), then there are no real solutions. This is true when the discriminant