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My name is Ann [436]
3 years ago
14

Annalise realized she needed two part-time jobs after school to raise enough money to buy a car within a month after getting her

driver’s license. She works the same number of hours at each job in one week and receives $10 per hour to babysit and $8 per hour to bag groceries. She also spends $11 per week on lunch at school. Which of the following equations represents the amount of money Annalise earns (y) in a week based on the number of hours ( x) she works?
y=18x+11


y=18x−11


y=10x+8


y=10x−3​
Mathematics
1 answer:
Lisa [10]3 years ago
3 0

Answer:

the correct answer is y=18x-11

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Help please!!!!!! And explain ur answer.. I will mark as BRAINLIEST​
Temka [501]

Answer:

Nu 4550

Step-by-step explanation:

Commission of 5% only applies to the first 70,000 so

70,000*\frac{5}{100} = 3500

That leaves 15,000 left in sales where the 7% commission applies.

15,000*\frac{7}{100}=1050

Adding these two together we get a total of

3500 +1050=Nu4550

7 0
3 years ago
Given the functions f (x) = x^2 + 4 and g (x) = 3x+ 8 what is a solution for the equation ? x^2 + 4 = 3x + 8 A. 0 B. 1 C. 2 D. 4
k0ka [10]

Answer:

D

Step-by-step explanation:

Let's solve your equation step-by-step.

x2+4=3x+8

Step 1: Subtract 3x+8 from both sides.

x2+4−(3x+8)=3x+8−(3x+8)

x2−3x−4=0

Step 2: Factor left side of equation.

(x+1)(x−4)=0

Step 3: Set factors equal to 0.

x+1=0 or x−4=0

x=−1 or x=4

Answer:

x=−1 or x=4

6 0
3 years ago
What is the value of a in this linear system?
Sergio [31]

Answer:

Step-by-step explanation:

C

5 0
3 years ago
Let an = –3an-1 + 10an-2 with initial conditions a1 = 29 and a2 = –47. a) Write the first 5 terms of the recurrence relation. b)
zlopas [31]

We can express the recurrence,

\begin{cases}a_1=29\\a_2=-47\\a_n=-3a_{n-1}+10a_{n-2}7\text{for }n\ge3\end{cases}

in matrix form as

\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}

By substitution,

\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}\implies\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^2\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}

and continuing in this way we would find that

\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}\begin{bmatrix}a_2\\a_1\end{bmatrix}

Diagonalizing the coefficient matrix gives us

\begin{bmatrix}-3&10\\1&0\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}

which makes taking the (n-2)-th power trivial:

\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}^{n-2}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}

\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}

So we have

\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}

and in particular,

a_n=\dfrac{29\left(2(-5)^{n-1}+5\cdot2^{n-1}\right)-47\left(-(-5)^{n-1}+2^{n-1}\right)}7

a_n=\dfrac{105(-5)^{n-1}+98\cdot2^{n-1}}7

a_n=15(-5)^{n-1}+14\cdot2^{n-1}

\boxed{a_n=-3(-5)^n+7\cdot2^n}

6 0
3 years ago
Please help me i'm struggling!
Marina CMI [18]

Answer:

y = \sqrt{x-5}

This is the correct answer because I got it right.

7 0
3 years ago
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