The graph of this system of equations is used to solve 4x^2-3x+6=2x^4-9x^3+2x

Mathematics

2 answers:

dlinn [17]2 years ago

7 0

To get the solution of the 2 equations, the graphs are plotted on a grid. Since we are told that, <span>4x^2-3x+6=2x^4-9x^3+2x, then solution set is found where the two graphs intersect.
Since we need both values of x and y, the solution set is represented by;
y coordinates of the intersection points and
x coordinates of the intersection points. </span>

saul85 [17]2 years ago

4 0

Answer:

The x-coordinates of the intersection points.

Step-by-step explanation:

to solve this equation, we want to find the values of x that make the equation true.

Graphing the related system of equations, we find the point of intersection for the two. This is the solution to the system of equations; the x-coordinate of this is the solution to the equation listed above.

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Dmitrij [34]

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8 0

2 years ago

what is the answer here 90 student went to the zoo, 3 had hamburger milk and cake; had 5 milk and hamburger; 10 had cake and mil

dangina [55]

Answer:

a) 37

b) 2

c) 17

d) 8

Step-by-step explanation:

90 Students went to the zoo. 3 had hamburger, milk and cake; 5 had milk and hamburger, 10 had cake and milk; 8 had cake and hamburger; 24 had hamburger; 38 had cake; 20 had milk. How many had a. nothing b. cake only c. milk only d. hamburger only

Solution:

Let h represent students that ate hamburger, m represent students that had milk and c represent students that had cake.

Given that:

n(h ∩ m ∩ c) = 3, n(m ∩ h) = 5, n(c ∩ m) = 10, n(c ∩ h) = 8, n(h) = 24, n(c) = 38, n(m) = 20

The number of students that had nothing = n(h ∪ m ∪ C)'

The number of students that had only milk = n(m ∩ h' ∩ C')

The number of students that had only cake = n(m' ∩ h' ∩ C)

The number of students that had only hamburger = n(m' ∩ h ∩ C')

a) n(m ∩ h' ∩ C') = n(m) - n(m ∩ h) - n(c ∩ m) - n(h ∩ m ∩ c) = 20 - 5 - 10 - 3 = 2

n(m' ∩ h ∩ C') = n(h) - n(m ∩ h) - n(c ∩ h) - n(h ∩ m ∩ c) = 24 - 5 - 8 - 3 = 8

n(m' ∩ h' ∩ C) = n(m) - n(m ∩ c) - n(c ∩ h) - n(h ∩ m ∩ c) = 38 - 10 - 8 - 3 = 17

n(m ∩ h' ∩ C') + n(m' ∩ h ∩ C') + n(m' ∩ h' ∩ C) + n(h ∪ m ∪ C)' + n(h ∩ m ∩ c) + n(m ∩ h) + n(c ∩ m) + n(c ∩ h) = 90