A student can take three subjects in 40 ways.
<u>SOLUTION:</u>
Given that, there are 4 different math courses, 5 different science courses, and 2 different history courses.
A student must take one of each, how many different ways can this be done?
Now, number ways to take math course = 4
Number of ways to take science course = 5
Number of ways to take history course = 2
So, now, total possible ways = product of possible ways for each course = 4 x 5 x 2 = 40 ways.
Hence, a student can take three subjects in 40 ways.
Answer:
PQ and QR are congruent.
Step-by-step explanation:
The length of PQ = sqrt [(2 - -1)^2 + (-1 - 3)^2]
= sqrt 25
= 5 units.
QR = sqrt [(5-2)^2 + (3 - -1)^2) ]
= sqrt 25
= 5 units.
PR = sqrt [ ( 3-3^2 + (5- -1)^2]
= sqrt 36
= 6 units.
Using the profit concept, it is found that she makes a profit of $4 selling a dozen cookies for six dollars.
<h3>What is a profit?</h3>
The profit is given by the <u>subtraction of the revenue by the total costs</u>.
In this problem:
- For the dozen cookies, she earned a revenue of 6 dollars.
- In all, the total cost of the 12 cookies was of 0.5 + 0.7 + 0.4 + 0.4 = 2 dollars.
6 - 2 = 4, hence she makes a profit of $4.
More can be learned about the profit concept at brainly.com/question/4001746
Answer:
Infinite Solutions
Step-by-step explanation:
x + 2y = 10
6y = 3x - 30
To solve for x and y we use substitution method
Let's solve the first equation for x
x + 2y = 10
Subtract 2y on both sides
x = 10 - 2y
Now plug in x in second equation
6y = -3x + -30
6y = -3 (10-2y) - 30
6y = -30 + 6y - 30
6y = 6y
Both sides are the same, so both x and y have infinite solutions.
It’s the third one!!! 1/2x-2
