1 cup of chocolate chips for every 1and 3/4 cup of flour
so divide 4 and 2/3 x 1 and 3/4
turn them both into fractions
4 2/3 = 14/3
1 3/4 = 7/4
14/3 / 7/4 = 14/3 x 4/7 = 56/21 = 2 and 2/3 cups of chocolate chips are needed
Answer:
(0, -3)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Algebra I</u>
- Terms/Coefficients
- Coordinates (x, y)
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
6x - 5y = 15
x = y + 3
<u>Step 2: Solve for </u><em><u>y</u></em>
<em>Substitution</em>
- Substitute in <em>x</em>: 6(y + 3) - 5y = 15
- Distribute 6: 6y + 18 - 5y = 15
- Combine like terms: y + 18 = 15
- [Subtraction Property of Equality] Subtract 18 on both sides: y = -3
<u>Step 3: Solve for </u><em><u>x</u></em>
- Define original equation: x = y + 3
- Substitute in <em>y</em>: x = -3 + 3
- Add: x = 0
Answer:
x=6
y=3
Step-by-step explanation:
y(side)= x
x(side)=2x
3rd side (3
) = x
x=3;
x= 2(3)=6
y=(3)=3
Answer:
1. ![(\sqrt[5]{(m+2)})^{3} = (m+2)^{\frac{3}{5}}](https://tex.z-dn.net/?f=%28%5Csqrt%5B5%5D%7B%28m%2B2%29%7D%29%5E%7B3%7D%20%3D%20%20%28m%2B2%29%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D)
2. ![(\sqrt[3]{(m+2)})^{5} = (m+2)^{\frac{5}{3}}](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7B%28m%2B2%29%7D%29%5E%7B5%7D%20%3D%20%20%28m%2B2%29%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D)
3. ![\sqrt[5]{(m)}^{3}+2 = m^{\frac{3}{5}}+2](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B%28m%29%7D%5E%7B3%7D%2B2%20%3D%20%20m%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D%2B2)
4. ![\sqrt[3]{(m)}^{5}+2 = m^{\frac{5}{3}}+2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%28m%29%7D%5E%7B5%7D%2B2%20%3D%20%20m%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%2B2)
Step-by-step explanation:
Recall that
![(\sqrt[n]{x})^{m} = (x^{\frac{m}{n}})](https://tex.z-dn.net/?f=%28%5Csqrt%5Bn%5D%7Bx%7D%29%5E%7Bm%7D%20%3D%20%20%28x%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%29)
Where
is called radicand and n is called index
1. Root(5, (m + 2) ^ 3)
In this case,
n is 5
m is 3
x = (m + 2)
![(\sqrt[5]{(m+2)})^{3} = (m+2)^{\frac{3}{5}}](https://tex.z-dn.net/?f=%28%5Csqrt%5B5%5D%7B%28m%2B2%29%7D%29%5E%7B3%7D%20%3D%20%20%28m%2B2%29%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D)
2. Root(3, (m + 2) ^ 5)
In this case,
n is 3
m is 5
x = (m + 2)
![(\sqrt[3]{(m+2)})^{5} = (m+2)^{\frac{5}{3}}](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7B%28m%2B2%29%7D%29%5E%7B5%7D%20%3D%20%20%28m%2B2%29%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D)
3. Root(5, m ^ 3) + 2
In this case,
n is 5
m is 3
x = m
![\sqrt[5]{(m)}^{3}+2 = m^{\frac{3}{5}}+2](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B%28m%29%7D%5E%7B3%7D%2B2%20%3D%20%20m%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D%2B2)
4. Root(3, m ^ 5) + 2
In this case,
n is 3
m is 5
x = m
![\sqrt[3]{(m)}^{5}+2 = m^{\frac{5}{3}}+2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%28m%29%7D%5E%7B5%7D%2B2%20%3D%20%20m%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%2B2)
Answer:
Step-by-step explanation:
The resultant can be calculated using triangular law of vector addition as :