Answer:
(x, y) = (40, 30)
Step-by-step explanation:
A graphing calculator can show you the solution to this system of equations is (x, y) = (40, 30). That is the point of intersection where the two lines cross.
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An algebraic solution can be found by using the substitution method. An expression for y can be found using the second equation:
y = 110 -2x . . . . . . subtract 2x from both sides
Using this in the first equation gives ...
3x -4(110 -2x) = 0 . . . . substitute for y
11x = 440 . . . . . . . . . simplify, add 440
x = 40 . . . . . . . . . . divide by 11
y = 110 -2(40) = 30
The solution is (x, y) = (40, 30).
|x| < 6 ⇔ x < 6 ∧ x > -6 ⇒ {x| x > -6 and x < 6}
Answer: b.
It already is in distributive property
1/ 816
Step-by-step explanation:
Step 1 :
Given
Total number of students in the school = 18
Number of students randomly selected = 3
Step 2 :
Number of ways in which 3 students can be selected out of 18 students is
C(18,3) = 18! /(3!(*18-3)!) = 18*17*16/3*2*1 = 816
Number of ways in which 3 students are selected and they are the youngest = 1
Hence the probability of selecting 3 youngest students from the available students is 1/ C(18,3) = 1/ 816
Imara used these steps to find the length of the hypotenuse of the right triangle.
3 squares are positioned to form a right triangle. The small square is labeled 15, medium square is 20, and large square is not labeled.
Step 1: Find the area of the square with side lengths of 20: 400
Step 2: Find the area of the square with side lengths of 15: 225
Step 3: Find the sum of the areas of the two squares: 625
Step 4: State the length of the hypotenuse: 625
Which best describes Imara’s error?
She should have found the area of the square with side lengths of 15 first.
She did not correctly calculate the area of the square with side lengths of 15.
She should have found the sum of 15 and 20 and then squared the sum.
She did not find the side lengths of the square with an area of 625.