Concept:
we use the vector method to solve this problem, take all the vertices in vector foam.
In the rectangle their exist two pair of parallel and equal sides, take the one pair and solve for unknown vertex.
BA=CD
A-B = D-C
(-2i-j)-(-2i-4j) = (xi+yj)-(1i-4j)
(-2i-j+2i+4j) = (xi+yj-1i+4j)
3j = xi+yj-1i+4j
i+3j-4j = xi+yj
1i-1j= xi+yj
by comparing,
x= 1
y= -1
so D vertex is (1, -1)
The correct answer
320
:)
2) Find an equivalent function to f(x) = 4(7)2x.
A) f(x) = 282x
B) f(x) = 4(49)x
C) f(x) = 196x
D) f(x) = 16x(49)x
3) The functions f(x) and g(x) are described using the following equation and table:
f(x) = −6(1.02)^x
x g(x)
−1 −5
0 −3
1 −1
2 1
Which equation best compares the y-intercepts of f(x) and g(x)?
a) The y-intercept of f(x) is equal to the y-intercept of g(x).
b) The y-intercept of f(x) is equal to 2 times the y-intercept of g(x).
c) The y-intercept of g(x) is equal to 2 times the y-intercept of f(x).
d) The y-intercept of g(x) is equal to 2 plus the y-intercept of f(x).
4) What is the value of x in the solution to the following system of equations? (5 points)
x − y = −3
x + 3y = 5
For #1 i put in 3 as the answer.
For #2 i put B as the answer.
For #3 i put B as the answer.
For #4 i put 2 as the answer.
Answer:
1) 3
2) B
3) B
4) x=-1
Step-by-step explanation:
1) The graph of is replaced by the graph of If g(x) is obtained by shifting f(x) down by 3 units, the value of k is 3.
2) Consider the function Note that Then
3) The y-intercept of f(x) is point (0,-6), because The y-intercept of g(x) is point (0,-3). Then the y-intercept of f(x) is equal to 2 times the y-intercept of g(x).
4) From the first equation Substitute it into the second equation:
Then
All of your answers are correct, except the last one.
The last one is a system of linear equations that you can solve by using elimination:
x - y = -3
x + 3y = 5
Subtract the two equations:
-4y = -8
y = 2
Use y to solve for x:
x - (2) = -3
Add 2 to both sides:
x = -1
The answer for #4 is x = -1.
A. P(Both defective) = start fraction six over 25 end fraction
B. P(Both defective) = start fraction one over 118 end fraction
C. P(Both defective) = start fraction three over 250 end fraction
D. P(Both defective) = start fraction nine over 625 end fraction
The probability that the two selected cans are defective, when the quality control worker selects the two cans from a batch of 60 cans is 1/118.
Probability of an event is the ratio of number of favorable outcome to the total number of outcome of that event.
A manufacturer produces soda cans and a quality control worker randomly selects two cans from the assembly line for testing.
Past statistics show that 10% of the cans are defective. Thus the probability of defective can to be selected is,
There is total 60 cans in which 10 percent are defective. Hence the total defective can in one batch are 6.
In the second attampt the number of total can in batch will be 59 and number of defective can will be 5.
The probability that the two selected cans are defective, when the quality control worker selects the two cans from a batch of 60 cans by chain rule is,
Thus, the probability that the two selected cans are defective, when the quality control worker selects the two cans from a batch of 60 cans is 1/118.
Learn more about the probability here;