Write 2x+y=17 in slope-intercept form. Solve for y?
1 answer:
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I'll explain it simply for you
1st question
Of course you know phythagoras theorm
You even wrote it up there
It states that the sum of the square of the two sides of an equilateral triangle is equal to the square of the hypotenuse
Where C is the hypotenuse
*NOTE* :
HYPOTENUSE is the greatest side in a triangle!!
And that's where your mistake is!
So you should take the greatest side as C
So in Q3. 7, 24 and 26 are the given numbers
You'll make the smaller two numbers a and b and the greatest number C
Using the Formula you'll solve the left side first which is
Then the right side which is
And if both are equal then it is a right triangle otherwise it isn't!
Let
a=7
b=24
c=26
a^2 + b^2
7^2 + 24^2
49 + 576 = 625
GREAT, Now the right side
26^2 = 676
Since they aren't equal it isn't a right angled triangle...
Then let
a=7.5
b=10
c=12.5
7.5^2 + 10^2
= 56.25 + 100
= 156.25
12.5^2 = 156.25
They are EQUAL
Therefore it is a right triangle too
Hopefully I helped
1st question
Of course you know phythagoras theorm
You even wrote it up there
It states that the sum of the square of the two sides of an equilateral triangle is equal to the square of the hypotenuse
Where C is the hypotenuse
*NOTE* :
HYPOTENUSE is the greatest side in a triangle!!
And that's where your mistake is!
So you should take the greatest side as C
So in Q3. 7, 24 and 26 are the given numbers
You'll make the smaller two numbers a and b and the greatest number C
Using the Formula you'll solve the left side first which is
Then the right side which is
And if both are equal then it is a right triangle otherwise it isn't!
Let
a=7
b=24
c=26
a^2 + b^2
7^2 + 24^2
49 + 576 = 625
GREAT, Now the right side
26^2 = 676
Since they aren't equal it isn't a right angled triangle...
Then let
a=7.5
b=10
c=12.5
7.5^2 + 10^2
= 56.25 + 100
= 156.25
12.5^2 = 156.25
They are EQUAL
Therefore it is a right triangle too
Hopefully I helped
The average travel time is simply the mean travel time between the given points
- The average time that it takes for the car to travel the first 0.25m is 2.23s.
- The average time to travel just between 0.25m and 0.50m is 0.9s.
- Given the time taken to travel the second 0.25 m section, the velocity would be 0.28m/s.
<u>(a) The average time to travel the first 0.25m</u>
The time travel in the first 0.25m are: 2.24s, 2.21s and 2.23s.
So, the average time to travel is:
<u>(b) The average time to travel just between 0.25m and 0.50m</u>
The time travel in the 0.50m are: 3.16s, 3.08s and 3.15s.
So, the average time to travel this distance is:
The average time to travel between both distance is the difference between the average time of each distance.
So, we have:
<u>(c) The velocity in the second 0.25m section</u>
The distance and time are:
So, the velocity is:
This gives
Read more about distance, velocity and time at:
