They bought 5 trophies total
1st trophy cost $37.50
($18.50+$12.50+$4+$2.50=$37.50)
Each additional trophy cost $15
($12.50+$2.50=$15)
They spent $97.50 total
$97.50-$37.50=$60
Additional trophies $15
$15*X=$60
4 trophies=$60+$37.50 for 1st trophy
$60+$37.50=$97.50
Problem 1
y + (y/a) = b
a*( y + (y/a) ) = a*b ... multiply both sides by 'a'
ay + a(y/a) = ab ... distribute
ay + y = ab
(a+1)y = ab .... factor
y(a+1) = ab
y = ab/(a+1) ... divide both sides by (a+1)
<h3>Answer: y = ab/(a+1)</h3>
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Problem 2
z - a = z/b
b(z - a) = b(z/b) .... multiply both sides by b
b(z - a) = z
bz - ab = z ... distribute
bz - ab+ab = z+ab ... add ab to both sides
bz = z+ab
bz-z = z+ab-z ... subtract z from both sides
bz-z = ab
z(b-1) = ab .... factor
z = ab/(b-1) .... divide both sides by (b-1)
<h3>Answer: z = ab/(b-1) </h3>
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Notes:
To clear out the fractions, I multiplied both sides by the denominator value.
The restriction that 'a' cannot equal -1, back in problem 1, is to avoid having the denominator (a+1) be equal to zero. We cannot divide by zero. A similar situation happens with problem 2 as well.
Cosine law
a^2 = b^2 + c^2 - 2(b)(c)cosA
a^2 = 2^2 + 4^2 - 2(2)(4)cos78
a = 4.08
a = 4.1
The STQ is obtuse i think that’s it?
Answer:
a. More than one pie
Step-by-step explanation:
2/3 + 3/4
8/12 + 9/12 = 17/12
17/12 = 1 5/12
