I believe the correct question is:
6.7*10^7 calc/sec?
So:
1/(6.7 x 10^7) = seconds per calculation
1/(6.7 x 10^7) = 100/(6.7 x 10^9) = 14.9 x 10^-9 seconds per
calculation = 14.9 ns
Which is approximately 15 ns
Answer:
15 ns
The first thing we must do in this case is find the derivatives:
y = a sin (x) + b cos (x)
y '= a cos (x) - b sin (x)
y '' = -a sin (x) - b cos (x)
Substituting the values:
(-a sin (x) - b cos (x)) + (a cos (x) - b sin (x)) - 7 (a sin (x) + b cos (x)) = sin (x)
We rewrite:
(-a sin (x) - b cos (x)) + (a cos (x) - b sin (x)) - 7 (a sin (x) + b cos (x)) = sin (x)
sin (x) * (- a-b-7a) + cos (x) * (- b + a-7b) = sin (x)
sin (x) * (- b-8a) + cos (x) * (a-8b) = sin (x)
From here we get the system:
-b-8a = 1
a-8b = 0
Whose solution is:
a = -8 / 65
b = -1 / 65
Answer:
constants a and b are:
a = -8 / 65
b = -1 / 65
Answer:
y = x - 2
Step-by-step explanation:
y = mx + b
5 = 1 (7) + b
5 = 7 + b
b = -2
So...
y = x - 2
Answer: 1
Step-by-step explanation:
8/-4 ÷ -3/9
8/-4 × 9/ -3 = 6
Answer:
Step-by-step explanation:
Vertex A of the triangle ABC when rotated by 90° counterclockwise about the origin,
Rule to be followed,
A(x, y) → P(-y, x)
Therefore, A(1, 1) → P(-1, 1)
Similarly, B(3, 2) → Q(-2, 3)
C(2, 5) → R(-5, 2)
Triangle given in second quadrant will be the triangle PQR.
If the point P of triangle PQR is reflected across a line y = x,
Rule to be followed,
P(x, y) → X(y, x)
P(-1, 1) → X(1, -1)
Similarly, Q(-2, 3) → Y(3, -2)
R(-5, 2) → Z(2, -5)
Therefore, triangle given in fourth quadrant is triangle XYZ.
