Answer:
Step-by-step explanation:
The change of base formula is
so filling in our given:
(Note: you do not have to put the base of 10 in the change of base formula because the "normal" base for a log is 10 and that is how your calculator figures it.)
Do this on your calculator to get
Not sure how many decimal places you needed!
Answer:
The temperature, in December is 47 °F
Step-by-step explanation:
The given data for the temperature are;
Month Temp (°F)
April 59
May 68
June 71
July 78
August 74
Let x represent the month, and f(x) = y represent the temperature, we have;
y = a·x² + b·x + c
When x = 1, we have, f(1) = a·1² + b·1 + c = 59
f(1) = a + b + c = 59...(1)
When x = 2, we have, f(2) = a·2² + b·2 + c = 68
f(2) = 4·a + 2·b + c = 68...(2)
When x = 3, we have, f(3) = a·3² + b·3 + c = 71
f(3) = 9·a + 3·b + c = 71...(3)
When x = 4, we have, f(4) = a·4² + b·4 + c = 78
f(3) = 16·a + 4·b + c = 78...(4)
When x = 5, we have, f(5) = a·5² + b·5 + c = 74
f(3) = 25·a + 5·b + c = 74...(5)
Subtracting equation (3) from equation (5), we get;
25·a + 5·b + c - (9·a + 3·b + c) = 74 - 71
16·a + 2·b = 3...(6)
Subtracting equation (1) from equation (2), we get;
4·a + 2·b + c - (a + b + c) = 68 - 59
3·a + b = 9...(7)
Multiply equation (7) by 2 and substract from equation (6) gives;
16·a + 2·b - 2 × (3·a + b) = 3 - 2 × 9
16·a + 2·b - 6·a - 2·b = 3 - 2 × 9
16·a - 6·a + 2·b - 2·b = 3 - 2 × 9 = -15
10·a = -15
a = -15/10 = -1.5
a = -1.5
From, 3·a + b = 9, we have;
3 × (-1.5) + b = 9
b = 9 + 4.5 =13.5
b = 13.5
From, a + b + c = 59, we have;
-1.5 + 13.5 + c = 59
c = 59 - (-1.5 + 13.5) = 47
c = 47
The quadratic equation becomes, y = a·x² + b·x + c = -1.5·x² + 13.5·x + 47
f(x) = y = -1.5·x² + 13.5·x + 47
For December, we have, x = 9, and f(x) = -1.5×9² + 13.5×9 + 47 = 47
The temperature, in December = 47 °F