The slope of the line is calculated using
y2 - y1 / x2 - x1
Substituting the given values
-8 - 27 / 5 - 0 = -7
The rate of change or the slope is -7
And the initial value is the value of y when x is 0. From the first coordinates, the initial value is 27.
Answer:
Step-by-step explanation:
The mnemonic SOH CAH TOA helps you remember the definitions of the trig functions. In particular, the function that involves the Opposite side and the Hypotenuse is the Sine function:
... Sin = Opposite/Hypotenuse
In this problem that means ...
... sin(55°) = x/(12 ft)
Multiplying by 12 ft, we have ...
... (12 ft)·sin(55°) = x ≈ 9.82982 ft ≈ 9.8 ft
Answer:
3216/8 is = to 402
Step-by-step explanation:
So I am not sure what you are asking sorry
Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).
