Answer:
x = 3/4 + (3 sqrt(5))/4 or x = 3/4 - (3 sqrt(5))/4
Step-by-step explanation:
Solve for x over the real numbers:
8 x^2 - 12 x - 23 = -5
Divide both sides by 8:
x^2 - (3 x)/2 - 23/8 = -5/8
Add 23/8 to both sides:
x^2 - (3 x)/2 = 9/4
Add 9/16 to both sides:
x^2 - (3 x)/2 + 9/16 = 45/16
Write the left hand side as a square:
(x - 3/4)^2 = 45/16
Take the square root of both sides:
x - 3/4 = (3 sqrt(5))/4 or x - 3/4 = -(3 sqrt(5))/4
Add 3/4 to both sides:
x = 3/4 + (3 sqrt(5))/4 or x - 3/4 = -(3 sqrt(5))/4
Add 3/4 to both sides:
Answer: x = 3/4 + (3 sqrt(5))/4 or x = 3/4 - (3 sqrt(5))/4
For each curve, plug in the given point
and check if the equality holds. For example:
(I) (2, 3) does lie on
since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.
For part (a), compute the derivative
, and evaluate it for the given point
. This is the slope of the tangent line at the point. For example:
(I) The derivative is

so the slope of the tangent at (2, 3) is

and its equation is then

For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents,
. For example:
(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation

5/9 and 11/21
LCM of the denominators is 63
changing their denominators to all have a common denominator of 63:
5(7)/63 = 35/63
11(3)/63 = 33/63
from the above;
5/9 > 11/21.
Answer:
5850
Step-by-step explanation:
is the question complete? if so this is the answer pls check if the question has more details
Answer:
(a)No. The probability of drawing a specific second card depends on the identity of the first card.
(b)4/663
(c) 4/663
(d) 8/663
Step-by-step explanation:
(a)The events are not independent because we are drawing cards without replacement and the probability of drawing a specific second card depends on the identity of the first card.
(b) P(ace on 1st card and jack on 2nd).

(c)P(jack on 1st card and ace on 2nd)

(d)Probability of drawing an ace and a jack in either order.
We can either draw an ace first, jack second or jack first, ace second.
Therefore:
P(drawing an ace and a jack in either order) =P(AJ)+(JA)
From parts (b) and (c) above:
