Answer:
B
Step-by-step explanation:
So far, we know that:
∠D = ∠J.
And that:
DE:JK = 14:7 = 2:1
So, to prove that ΔDEF ~ ΔJKL by SAS, DF must be similar to JL, as those are the sides between the angle.
So:
DF:JL = 2:1.
Our answer is B.
Answer:
see explanation
Step-by-step explanation:
note when x = - 3
(- 3)³ + 2(- 3)² + 4(- 3) + 21 = - 27 + 18 - 12 + 21 = 0
hence x = - 3 is a zero and (x + 3) is a factor and dividing gives
= (x + 3)(x² - x + 7)
For zeros equate to zero
(x + 3)(x² - x + 7) = 0
equate each factor to zero and solve for x
x + 3 = 0 ⇒ x = - 3
x² - x + 7 = 0 ← solve using quadratic formula
x = (1 ±
) / 2 = (1 ± 3i
) / 2
x =
± 
zeros are x = - 3, x =
±
Answer:
e.none of these
Step-by-step explanation:
Computations For CC for Fraction defective
Sample No d p=d/100
1 0 0
2 0 0
3 2 0.02
4 1 0.01
5 0 0
6 1 0.01
7 2 0.02
8 0 0
Total 0.06


3 sigma control limits for p chart are given by:


hence option e is correct
The equation that you are looking for is y = -2x + 3
Same as before
total=(2w+75)(2w+35)=4w²+220w+2625
pool=75 times 35=2625
minus pool from total
4w²+220w+2625-2625=4w²+220w