If we know your Pythagorean Triples we can immediately recognize that the last choice is a right triangle:
8² + 15² = 17²
If you don't know your Pythagorean Triples, it's worth learning the first few off the list because teachers use them in problems all the time. But for now let's just exhaustively check the Pythagorean Theorem for each triangle. We don't have to multiply everything out; we can analyze the common factors. If two have a common factor that the third one doesn't have, there's no way for the Pythagorean Theorem to add up.
Clearly 5²+15² is a multiple of 5 but 18² isn't so that one isn't a right triangle.
6²+12² is a multiple of 6, 16² isn't a multiple of 6, not an RT.
15²-5² is a multiple of 5, 13² isn't, no joy.
8²+15² = 64 + 225 = 289 = 17² -- that's a real right triangle, a valid Pythagorean Triple.
Answer:
Step-by-step explanation:
Rent, wages and car insurance don't vary much. Grocery expenses depend upon the particular groceries purchased each time one goes to the grocery store, and are least likely to be steady / constant from week to week.
3x - 3y + 9 = 0
The y-intercept is the point on the graph where it crosses the y-axis, and has coordinates of (0, b). It is also the value of y when x = 0.
To solve for the y-intercept, set x = 0:
3(0) - 3y + 9 = 0
3(0) - 3y + 9 = 0
Subtract 9 from both sides:
- 3y + 9 - 9 = 0 - 9
- 3y = -9
Divide both sides by -3 to solve for y:
-3y/-3 = -9/-3
y = 3
Therefore, the y-intercept is (0, 3).
The x-intercept is the point on the graph where it crosses the x-axis, and has coordinates of (a, 0). It is also the value of x when y = 0.
To solve for the x-intercept, set y = 0:
3x - 3(0)+ 9 = 0
3x -0 + 9 = 0
Subtract 9 from both sides:
3x + 9 - 9 = 0 - 9
3x = -9
Divide both sides by 3 to solve for x:
3x/3 = -9/3
x = -3
Therefore, the x-intercept is (-3,0).
The correct answers are:
Y-intercept = (0, 3)
X-intercept = (-3, 0)
Answer:
Step-by-step explanation:
for c it would be 1\8
for E it would be 3\8
Let's to the first example:
f(x) = x^2 + 9x + 20
Ussing the formula of basckara
a = 1
b = 9
c = 20
Delta = b^2 - 4ac
Delta = 9^2 - 4.(1).(20)
Delta = 81 - 80
Delta = 1
x = [ -b +/- √(Delta) ]/2a
Replacing the data:
x = [ -9 +/- √1 ]/2
x' = (-9 -1)/2 <=> - 5
Or
x" = (-9+1)/2 <=> - 4
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Already the second example:
f(x) = x^2 -4x -60
Ussing the formula of basckara again
a = 1
b = -4
c = -60
Delta = b^2 -4ac
Delta = (-4)^2 -4.(1).(-60)
Delta = 16 + 240
Delta = 256
Then, following:
x = [ -b +/- √(Delta)]/2a
Replacing the information
x = [ -(-4) +/- √256 ]/2
x = [ 4 +/- 16]/2
x' = (4-16)/2 <=> -6
Or
x" = (4+16)/2 <=> 10
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Now we are going to the 3 example
x^2 + 24 = 14x
Isolating 14x , but changing the sinal positive to negative
x^2 - 14x + 24 = 0
Now we can to apply the formula of basckara
a = 1
b = -14
c = 24
Delta = b^2 -4ac
Delta = (-14)^2 -4.(1).(24)
Delta = 196 - 96
Delta = 100
Then we stayed with:
x = [ -b +/- √Delta ]/2a
x = [ -(-14) +/- √100 ]/2
We wiil have two possibilities
x' = ( 14 -10)/2 <=> 2
Or
x" = (14 +10)/2 <=> 12
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To the last example will be the same thing.
f(x) = x^2 - x -72
a = 1
b = -1
c = -72
Delta = b^2 -4ac
Delta = (-1)^2 -4(1).(-72)
Delta = 1 + 288
Delta = 289
Then we are going to stay:
x = [ -b +/- √Delta]/2a
x = [ -(-1) +/- √289]/2
x = ( 1 +/- 17)/2
We will have two roots
That's :
x = (1 - 17)/2 <=> -8
Or
x = (1+17)/2 <=> 9
Well, this would be your answers.
