Answer:
You need to find the a factor of 35 that when added will result to 16. and then you just have to square each number and add them...
I tried solving but all the factors of 35 when added is not equal to 16 but one equals to 12 only..
hope I'm right
Marcos has 6 nickels and 9 quarters
<em><u>Solution:</u></em>
Let "n" be the number of nickels
Let "q" be the number of quarters
<em><u>Marcos had 15 coins in nickels and quarters</u></em>
Therefore, we can say,
number of nickels + number of quarters = 15
n + q = 15 -------- eqn 1
<em><u>He has three more quarters than nickels</u></em>
Number of quarters = 3 + number of nickels
q = 3 + n -------- eqn 2
Eqn 1 and eqn 2 represents the system of equations
<em><u>Substitute eqn 2 in eqn 1</u></em>
n + 3 + n = 15
2n + 3 = 15
2n = 15 - 3
2n = 12
<h3>n = 6</h3>
<em><u>Substitute n = 6 in eqn 2</u></em>
q = 3 + 6
<h3>q = 9</h3>
Thus Marcos has 6 nickels and 9 quarters
Answer:
95%
Step-by-step explanation:
Step 1: 38/40= 0.95
Step 2: multiply 0.95x100= 95
Answer:
![x^2(x - 2)(x + 2)(x^2 + 4)](https://tex.z-dn.net/?f=x%5E2%28x%20-%202%29%28x%20%2B%202%29%28x%5E2%20%2B%204%29)
Step-by-step explanation:
![x^6-16x^2](https://tex.z-dn.net/?f=x%5E6-16x%5E2)
![x^2\left(x^4-16\right)](https://tex.z-dn.net/?f=x%5E2%5Cleft%28x%5E4-16%5Cright%29)
![x^2\left(x^2+4\right)\left(x+2\right)\left(x-2\right)](https://tex.z-dn.net/?f=x%5E2%5Cleft%28x%5E2%2B4%5Cright%29%5Cleft%28x%2B2%5Cright%29%5Cleft%28x-2%5Cright%29)
![\boxed{\bold{HOPE \:IT \:HELPS!\: :)}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cbold%7BHOPE%20%5C%3AIT%20%5C%3AHELPS%21%5C%3A%20%3A%29%7D%7D)
I use math-a-way it answers questions like that