Given a1 = 90
Common difference = adding
90 + 9 = 99, second term
99 + 9 = 108, third term
And keep doing that
You would get
90, 99, 108, 117, 126, 135
Answer:
a) The variable, M represents the amount of fish Ms. Montesinos has
b) The set equation is 11 + 2 = 3 × M - 2
c) 5
Step-by-step explanation:
The given parameters are;
The number of fishes in Ms. Cruz's tank = 11 fishes
The number of fishes Ms, Cruz has after buying two more fishes = 3 × The amount of fish Ms. Montesinos has - 2 fishes
a) Let the variable, M represent the amount of fish Ms. Montesinos has
b) The set up equation is as follows;
11 + 2 = 3 × M - 2
c) From the equation, 11 + 2 = 3 × M - 2, we have;
13 = 3·M - 2
3·M = 13 + 2 = 15
3·M = 15
M = 15/3 = 5
M = 5
The number of fishes Ms. Montesinos has is 17 fishes
Answer:
It's false
All quadrilaterals are also rhombus, <u>False</u>
18 = x
__ __
5 4
x = 18•4 / 5
72/5 = 14.4 final answer
