Question

1. a math test has 25 problems. some are worth 2 points, and some are worth 3 points. the test is worth 60 points total. which system can be used to determine the number of 2 - point problems and the number of 3 - point problems on the test? a (\begin{cases}x + y = 25\\2x + 3y = 60end{cases}) b (\begin{cases}x + y = 60\\2x + 3y = 25end{cases}) c (\begin{cases}x - y = 25\\2x + 3y = 60end{cases}) d (\begin{cases}x - y = 60\\2x - 3y = 25end{cases}) 34. an electrician charges $15 plus $11 per hour. another electrician charges $10 plus $15 per hour. for what amount of time will the cost be the same? what is that cost? f 1 hour; $25 g (1\frac{1}{4}) hours; $28.75 h (1\frac{1}{2}) hours; $30 j (1\frac{3}{4}) hours; $32.50

Explanation:

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Problem 33

Step1: Define variables

Let $x$ = number of 2-point problems, $y$ = number of 3-point problems.

Step2: Total problems equation

Total problems = 25: $x + y = 25$

Step3: Total points equation

Total points = 60: $2x + 3y = 60$

Step1: Define variables and cost functions

Let $t$ = time in hours.
Cost of 1st electrician: $C_1 = 15 + 11t$
Cost of 2nd electrician: $C_2 = 10 + 15t$

Step2: Set costs equal

Find $t$ when $C_1 = C_2$:
$15 + 11t = 10 + 15t$

Step3: Solve for $t$

Rearrange terms: $15 - 10 = 15t - 11t$
$5 = 4t$
$t = \frac{5}{4} = 1\frac{1}{4}$

Step4: Calculate total cost

Substitute $t=1\frac{1}{4}$ into $C_1$:
$C_1 = 15 + 11\times\frac{5}{4} = 15 + \frac{55}{4} = 15 + 13.75 = 28.75$

Answer:

A.

$$\begin{cases} x + y = 25 \\ 2x + 3y = 60 \end{cases}$$

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Problem 34