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3 years ago

If cos Theta < 0, which quadrent(s) could the terminal side of Theta lie?

Mathematics

1 answer:

Korolek [52]3 years ago

4 0

If cosΦ is lesser than zero (negative) it must be in quadrant 2 or 3 because cos is always positive in quadrants 1,4.

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The sum of a number and twice its reciprocal is 27/5. Find the number.

vovangra [49]

Answer:

x , reciprocal= 1/x

x+2(1/x)=27/5

x+2/x=27/5

x^2+2=27/5*x

x^2-27/5*x+2=0

(x-5)(x-2/5)=0

x=5, 2/5

6 0

3 years ago

2. When Joey dives off a diving board, the equation of his pathway can be modeled by h = -16- + 15 + 12.

Sergeu [11.5K]

Answer:

a) The maximum height is approx 15.5 unit.

b) The time it will take for Joey to reach the water is 1.45 hour.

Step-by-step explanation:

Given : When Joey dives off a diving board, the equation of his pathway can be modeled by $h(t)= -16t^2+15t + 12$

To find : a) Find Joey's maximum height.

b) Find the time it will take for Joey to reach the water.

Solution :

Modeled $h(t)= -16t^2+15t + 12$ ....(1)

a) To find maximum height

Derivate (1) w.r.t. t,

$h'= -32t+15$

For critical point put h'=0,

$-32t+15=0$

$t=\frac{15}{32}$

$t=0.46875$

Derivate again w.r.t. t,

$h''= -32$

It is maximum at t=0.46875.

Substitute t in equation (1),

$h(0.46875)= -16(0.46875)^2+15(0.46875)+12$

$h(0.46875)= -3.515625+7.03125+12$

$h(0.46875)= 15.515625$

The maximum height is approx 15.5 unit.

b) To find the time it will take for Joey to reach the water.

Put h=0 in equation (1),

$-16t^2+15t + 12=0$

Apply quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, a=-16 , b=15, c=12

$t=\frac{-15\pm\sqrt{(15)^2-4(-16)(12)}}{2(-16)}$

$t=\frac{-15\pm\sqrt{225+768}}{-32}$

$t=\frac{-15\pm\sqrt{993}}{-32}$

$t=\frac{-15+\sqrt{993}}{-32},\frac{-15-\sqrt{993}}{-32}$

$t=−0.515,1.453$

Reject negative value.

The time is t=1.45.

The time it will take for Joey to reach the water is 1.45 hour.

5 0

3 years ago

Please help me find the answer! solve for x.

liq [111]

Answer:

answer is 72 degree because the side of the angle is increased from side AC

6 0

2 years ago

How much can 2 go into 80

iren [92.7K]

Answer:

Step-by-step explanation:

80/2= 40

8 0

2 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

Alexxandr [17]

Answer:

P = 0,0012 or P = 0.12 %

Step-by-step explanation:

We know for normal distribution that:

μ ± σ in that range we find 68.3 % of all values

μ ± 2σ ⇒ 95.5 % and

μ ± 3σ ⇒ 99.7 %

Fom problem statement

We have to find (approximately) % of cars that reamain in service between 71 and 83 months

65 + 6 = 71 ( μ + σ ) therefore 95.5 % of values are from 59 and up to 71 then by symmetry 95.5/2 = 49.75 of values will be above mean

Probability between 65 and 71 is 49.75 %

On the other hand 74 is a value for mean plus 1, 5 σ and

74 is the value limit for mean plus 1,5 σ and correspond to 49,85 (from z=0 or mean 65).

Then the pobabilty for 83 have to be bigger than 49.85 and smaller than 0,5 assume is 49.87

Finally the probability approximately for cars that remain in service between 71 and 83 months is : 0,4987 - 0.49.75

P = 0,0012 or P = 0.12 %

8 0

3 years ago