Well she charged twice as much to walk large then small. Let s be small and l be large.
We we know that
l = 2s
we also know that she wants to make 100 so
xl + ys = 100
Where where x is number of large and y is number of small. it says she walked 10 small and 5 large so
5l + 10s = 100
substitute 2s for l because they are same value as shown in first equation I made
5 (2s) + 10s = 100
10s + 10s = 100
20s = 100
s = 5
we can now solve for l
l = 2s
l = 2 * 5
l = 10
So s is 5 and l is 10.
Answer:
x = -203/23
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>
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>
y = -23(x + 9) + 4
y = 0
<u>Step 2: Solve for </u><em><u>x</u></em>
- Substitute in <em>y</em>: 0 = -23(x + 9) + 4
- [Subtraction Property of Equality] Subtract 4 on both sides: -4 = -23(x + 9)
- [Division Property of Equality] Divide -23 on both sides: 4/23 = x + 9
- [Subtraction Property of Equality] Subtract 9 on both sides: -203/23 = x
- Rewrite: x = -203/23
Answer:
about 5 boxes
Step-by-step explanation:
2 *40/15 Please give brainliest
Answer:
<em>f(x)=x²-3x-10</em>
Step-by-step explanation:
\begin{gathered}f(x) = x {}^{2} - 3x - 10 \\ to \: find \: x \: intercept \:o r \: zero \: substitute \: f(x) = 0\: \\ 0 = x {}^{2} - 3x - 10 \\ x {}^{2} - 3x - 10 = 0 \\ x {}^{2} + 2x - 5x - 10 = 0 \\ x(x + 2) - 5x - 10 = 0 \\ x(x + 2) - 5(x + 2) = 0 \\ (x + 2).(x - 5) = 0 \\ x + 2 = 0 \\ x - 5 = 0 \\ x = - 2 \\ x = 5\end{gathered}
f(x)=x
2
−3x−10
tofindxinterceptorzerosubstitutef(x)=0
0=x
2
−3x−10
x
2
−3x−10=0
x
2
+2x−5x−10=0
x(x+2)−5x−10=0
x(x+2)−5(x+2)=0
(x+2).(x−5)=0
x+2=0
x−5=0
x=−2
x=5
therefore the zeros of the equation are x₁=-2,x₂=5
