Step-by-step explanation:
(a×b^-3×c^-3)⁵ / (a^-3×b⁴×c⁴)^-5
let's look at the numerator (the top) first :
a⁵×b^-15×c^-15
and now the denominator (the bottom) :
a¹⁵×b^-20×c^-20
both divided are (due to the commutative rules of multiplication we can split this first into the parts of the individual variables, and then multiply them all with each other) :
a⁵/a¹⁵ = 1/a¹⁰
b^-15 / b^-20 = b^(-15 ‐ -20) = b⁵
c^-15 / c^-20 = c⁵
so we get as result :
b⁵c⁵/a¹⁰
Answer:
Step-by-step explanation:
Yes, she can use this inequality and it does matter since the number of cars that the inequality provides will need to be equal to or more than that number in order for all the students to be able to go. Therefore, if we apply the inequality it would give us the minimum number of cars needed (n) like so
12 + 3n > 28 ... subtract 12 on both sides
3n > 16 ... divide both sides by 3
n > 5 1/3
Since there can't be 1/3 of a car and the number of cars needed must be higher than 5 1/3 then we would need a total of 6 cars to take all of the children.
Answer:
x = 6
Step-by-step explanation:
These two angles make a complementary angle meaning the sum of the angles will be 90°. We can use this information to make an equation to solve for x.
10x + 30 = 90 (Given)
10x = 60 (Subtract 30 on both sides)
x = 6 (Divide 10 on both sides)
Answer:
Isn't the sum when you add numbers together? if this is true, there is no way you can get the sum of 28
Step-by-step explanation:
15,17,19
Suppose the three odd integers are #n#, #n+2# and #n+4#
Their sum is:
#n + (n+2) + (n+4) = 3n+6#
#13# more than twice the largest of the three is:
#2(n+4)+13 = 2n+21#
From what we are told these two are equal:
#3n+6 = 2n+21#
Subtract #2n+6# from both sides to get:
#n = 15#
So the three integers are:
#15, 17, 19#
