<span>A)Both m(x) and p(x) cross the x-axis at 7.
B)Both m(x) and p(x) cross the y-axis at 7.
C)Both m(x) and p(x) have the same output value at x = 7.
D)Both m(x) and p(x) have a maximum or minimum value at x = 7.
m(x) = p(x) at x = 7
</span><span>
True statement about x = 7.
C)Both m(x) and p(x) have the same output value at x = 7. </span><span>
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Answer:
Step-by-step explanation:
Problem One (left panel)
<em><u>Question A</u></em>
- The y intercept happens when x = 0
- That being said, the y intercept is 50. It was moving when the timing began.
<em><u>Question B</u></em>
The rate of change = (56 - 52)/(3 - 1) = 4/2 = 2 miles / hour^2 (you have a slight acceleration.
<em><u>Question C</u></em>
- 60 = a + (n-1)d
- 60 = 50 + (n - 1)*2
- 10/2 = (n - 1)*2/2
- 5 = n - 1
- 6 = n
The way I have done it the domain is n from 1 to 6
Question 2 (Right Panel)
<em><u>Question A</u></em>
The equation for the table is f(x) = 3x - 3 which was derived simply by putting all three points into y = ax + b and solving.
- f(0) = ax + b
- -3 = a*0) + b
- b = - 3
- So far what you have is
- f(x) = ax - 3
- f(-1) = a*(-1) - 3 but we know (f(-1)) = -6
- - 6 = a(-1) - 3 add 3 to both sides
- -6 +3 = a(-1) -3 + 3
- -3 = a*(-1) Divide by - 1
- a = 3
- f(x) = 3x - 3 Answer for f(x)
- The slope of f(x) = the coefficient in front of the x
- f(x) has a slope of 3
- g(x) has a slope of 4
<em><u>Part B</u></em>
- f(x) has a y intercept of - 3
- g(x) has a y intercept of -5
- f(x) has the greater y intercept.
- -3 > - 5
Answer:
1. 8x+9
2. 3x + 11
3. 9x + 8
4. 11x+9
5. 6x
6. 10x+ 8
7. 7x+7
Step-by-step explanation:
In expanded form that would be 10,000,000 + 6,000,000 + 100,000 + 7,000 + 300 + 20
Answer: a) An = An-1 + An-2
b) 55ways
Step-by-step explanation:
a) a nickel is 5 cents and a dime is 10cent so a multiple of 5 cents is the possible way to pay the tolls in both choices.
Let An represents the number of possible ways the driver can pay a toll of 5n cents, so that
An = 5n cents
Case 1: Using a nickel for payment which is 5 cents, the number of ways given as;
An-1 = 5( n-1)
Case 2: using a dime which is two 5 cents, the number of ways is given as;
An-2 = 5(n-2)
Summing up the number of ways, we have
An = An-1 + An-2
From the relation,
If n= 0, Ao= 1
n =1, A1= 1
b) 45 cents paid in multiples of 5cents will give us 9 ways(A9)
From the relation, we have that
Ao = 1
A1 = 1
An =An-1 + An-2
Ao = 1
A1 = 1
A2 = A1+Ao = 1+1= 2
A3 = A2 + A1 = 3
A4 = A3+A2=5
A5=A4+A3=8
A6=A5+A4=13
A7 =A6+A5 = 21
A8= A7+A6= 34
A9= A8+A7= 55
So there are 55ways to pay 45cents.
