The perpendicular line will have an equation of y=3/4x-3/4
To find this, we first have to solve our equation for slope intercept form.
8x + 6y = -5 ----> subtract 8x
6y = -8x + -5 ----> divide by 6
y = -4/3x - 5/6.
So we know the slope of this equation to be -4/3. Since perpendicular lines have opposite and reciprocal slopes, we know we can simply flip the fraction and make it a negative to get the new slope of 3/4. Since B is the only option with that slope, we know it to be the correct answer.
Answer:
The correct answer is option B. 17
Step-by-step explanation:
It is given that, ZX bisects ∠WZY. If the measure of ∠YXZ is (6m – 12)°
<u>To find the value of m</u>
From the figure we can see that, triangle WYZ is an isosceles triangle.
ZW = ZY
Then <YXZ = <WXZ = 90°
It is given ∠YXZ = (6m – 12)°
(6m – 12)° = 90°
6m = 90 + 12 = 102
m = 102/6 = 17
Therefore the value of m = 17
The correct answer is option B. 17
Answer: what the heck do you have to answer
Step-by-step explanation:
Answer:
- number of multiplies is n!
- n=10, 3.6 ms
- n=15, 21.8 min
- n=20, 77.09 yr
- n=25, 4.9×10^8 yr
Step-by-step explanation:
Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...
mpy[n] = n·mp[n-1]
mpy[2] = 2
So, ...
mpy[n] = n! . . . n ≥ 2
__
If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...
10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10
Then the larger matrices take ...
n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min
n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years
n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years
_____
For the shorter time periods (less than 100 years), we use 365.25 days per year.
For the longer time periods (more than 400 years), we use 365.2425 days per year.
