Answer:
x = -2bm/(m²+1)
Step-by-step explanation:
One line has equation
... y = mx + b
The slope of the perpendicular line is the negative reciprocal of the slope of the original line. The perpendicular line with the opposite y-intercept has equation
... y = (-1/m)x - b
The point of intersection is where the x- and y-values are equal, so ...
... mx + b = (-1/m)x - b
... (m +1/m)x = -2b . . . . . . . add 1/m - b to both sides
... (m²+1)x = -2bm . . . . . . . multiply by m
... x = -2bm/(m²+1) . . . . . . divide by the coefficient of x
Frank = F
Sue = S
John = J
F=3*S
F = J+15
S = J-1
If you want to find Frank's age, then his age would be equivalent to John's plus 15 years.
A.-Would not work because Frank is three times Sue's age, not John's (left hand side of the equation).
B.-Notice that the right hand side of the equation is equivalent to Sue's age, which we know is John-1, however it is currently written to be "three times Sue's age minus one" which would give us John's age, plus two more years than his actual age on the left hand side.
C.-Frank's age is equal to John's plus fifteen (right side of the equation) and Frank is equal to Sue's age times 3. But, if Sue is in terms of Johns, then Sue's age is John's minus one. Therefore, Frank's age is equal to three times Sue's age of John minus one, which is the left-hand side of our equation.
Therefore C is the answer. C:
Answer:
h=16
Step-by-step explanation:
h-9=7
h=9+7
h=16
I hope this helps.
First, tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>), so if cos(<em>θ</em>) = 3/5 > 0 and tan(<em>θ</em>) < 0, then it follows that sin(<em>θ</em>) < 0.
Recall the Pythagorean identity:
sin²(<em>θ</em>) + cos²(<em>θ</em>) = 1
Then
sin(<em>θ</em>) = -√(1 - cos²(<em>θ</em>)) = -4/5
and so
tan(<em>θ</em>) = (-4/5) / (3/5) = -4/3
The remaining trig ratios are just reciprocals of the ones found already:
sec(<em>θ</em>) = 1/cos(<em>θ</em>) = 5/3
csc(<em>θ</em>) = 1/sin(<em>θ</em>) = -5/4
cot(<em>θ</em>) = 1/tan(<em>θ</em>) = -3/4
Answer:
The lengths of Gajge’s runs have greater variability because there is a greater difference between his longest and shortest runs is the answer.
Step-by-step explanation:
given that Ty and Gajge are football players.
Carries is 15 for both and average is the same 4 for both.
But on scrutiny we find that maximum and minimum and 6 and 2 for Ty.
Hence range for Ty = 6-2 =4 (2 runs on eithre side of mean)
But for Gajge, highest is 19 and lowest is 2.
i.e. range = 19-2 =17 very much higher than that of Ty
The lengths of Gajge’s runs have greater variability because there is a greater difference between his longest and shortest runs.
