Using a system of equations, it is found that there are 5 dimes and 9 quarters in his pocket.
<h3>What is a system of equations?</h3>
A system of equations is when two or more variables are related, and equations are built to find the values of each variable.
In this problem, the variables are given as follows:
- Variable x: number of dimes in his pocket.
- Variable y: number of quarters in his pocket.
He has a total of 14 coins, hence:
x + y = 14 -> y = 14 - x.
They are worth $2.75, hence, considering the value of each coin(dimes $0.1 and quarters $0.25), we have that:
0.1x + 0.25y = 2.75
Since y = 14 - x:
0.1x + 0.25(14 - x) = 2.75
x = (0.25*14 - 2.75)/0.15.
x = 5.
y = 14 - x = 14 - 5 = 9.
There are 5 dimes and 9 quarters in his pocket.
More can be learned about a system of equations at brainly.com/question/24342899
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Answer:
1. The volume of the pile is 4,399.7719 m^3
2. The number of sanders to be filled from a pile is 638
Step-by-step explanation:
Here, we firstly would calculate the volume of the pile
Mathematically, that will be;
V = 1/3 * pi * r^2 * h
from the question, h = 14.2
Mathematically, r = d/2
r = 34.4/2 = 17.2 m
So the volume will be:
V = 1/3 * 3.142 * 17.2^2 * 14.2
V = 4,399.7719 m^3
To find the number of sanders to be filled,
we simply divide the volume obtained by the volume a sander can take
That will be;
4,399.7719/6.9
= 637.6 which is 638
Answer:
9/10
Step-by-step explanation:
Combined usually means to add. You only add the top.
(p + q)⁵
(p + q)(p + q)(p + q)(p + q)(p + q)
{[p(p + q) + q(p + q)][p(p + q) + q(p + q)](p + q)}
{[p(p) + p(q) + q(p) + q(q)][p(p) + p(q) + q(p) + q(q)](p + q)}
(p² + pq + pq + q²)(p² + pq + pq + q²)(p + q)
(p² + 2pq + q²)(p² + 2pq + q²)(p + q)
{[p²(p² + 2pq + q²) + 2pq(p² + 2pq + q²) + q²(p² + 2pq + q²)](p + q)}
{[p²(p²) + p²(2pq) + p²(q²) + 2pq(p²) + 2pq(2pq) + 2pq(q²) + q²(p²) + q²(2pq) + q²(q²)](p + q)}
(p⁴ + 2p³q + p²q² + 2p³q + 4p²q² + 2pq³ + p²q² + 2pq³ + q⁴)(p + q)
(p⁴ + 2p³q + 2p³q + p²q² + 4p²q² + p²q² + 2pq³ + 2pq³ + q⁴)(p + q)
(p⁴ + 4p³q + 6p²q² + 4pq³ + q⁴)(p + q)
p⁴(p + q) + 4p³q(p + q) + 6p²q²(p + q) + 4pq³(p + q) + q⁴(p + q)
p⁴(p)+ p⁴(q) + 4p³q(p) + 4p³q(q) + 6p²q²(p) + 6p²q²(q) + 4pq³(p) + 4pq³(q) + q⁴(p) + q⁴(q)
p⁵ + p⁴q + 4p⁴q + 4p³q² + 6p³q² + 6p²q³ + 4p²q³ + 4pq⁴ + pq⁴ + q⁵
p⁵ + 5p⁴q + 10p³q² + 10p²q³ + 5pq⁴ + q⁵
