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Katyanochek1 [597]
2 years ago
9

If T is the point at the given distance on the unit circle C from P (1, 0), determine the quadrant in which T lies:

Mathematics
2 answers:
givi [52]2 years ago
8 0
Hello,

Answer D

9π/6<11π/6<12π/6
Quadrant IV

stealth61 [152]2 years ago
6 0

Answer:

Option d) is correct

Step-by-step explanation:

T is a point at a given distance on the unit circle from P(0,1) .

Clearly, the point P(0,1) lies on the circle itself .

To find: the quadrant in which T lies

Given: angle at which point T lies is \frac{11\pi}{6}.

Solution :

\frac{11\pi}{6}=\frac{11\times 180^{\circ}}{6}=11\times 30^{\circ}=330^{\circ}

As 270^{\circ}< 330^{\circ}< 360^{\circ},

So, basically we are talking about the fourth quadrant as angle 330^{\circ} lies in the fourth quadrant .

Therefore, point T lies in quadrant IV.

i.e If T is the point at the given distance on the unit circle C from P (1, 0), then T lies in the IV quadrant.

So, option d) is correct .

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Answer: 12+8x

Step-by-step explanation:

1) 4(3+2x)- Distribute the 4

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What is the answer please.... need help
mina [271]

Answer:

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

f\left(x\right)=x^3-6x^2+3x+10

As the highest power of the x-variable is 3 with the leading coefficients of 1.

  • So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

f\left(x\right)=x^3-6x^2+3x+10

0=x^3-6x^2+3x+10              ∵  f(x)=0

as

Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0

so

\left(x+1\right)\left(x-2\right)\left(x-5\right)=0    

Using the zero factor principle

if  ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x=-1,\:x=2,\:x=5

Therefore, the zeros of the function are:

x=-1,\:x=2,\:x=5

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.    

8 0
2 years ago
hi um can you tell me where i am supposed to put the 2 points? i know that the first one goes on the 0 but where do i but the 2n
Nina [5.8K]

Answer:

114.75

Step-by-step explanation:

You just multiply 25.50 by 4.50 :))

Hope this helps and sorry if I'm wrong.

Have a wonderful rest of your day. <3

8 0
2 years ago
A rock is thrown upward from a bridge that is 57 feet above a road. The rock reaches its maximum height above the road 0.76 seco
bekas [8.4K]

answer:

f(t) = -9.9788169(t -0.76)^2 +57

Step-by-step explanation:

On this question we see that we are given two points on a certain graph that has a maximum point at 57 feet and in 0.76 seconds after it is thrown, we know can say this point is a turning point of a graph of the rock that is thrown as we are told that the function f determines the rocks height above the road (in feet) in terms of the  number of seconds t since the rock was thrown therefore this turning point coordinate can be written as (0.76, 57) as we are told the height represents y and x is represented by time in seconds. We are further given another point on the graph where the height is now 0 feet on the road then at this point its after 3.15 seconds in which the rock is thrown in therefore this coordinate is (3.15,0).

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0 = a(3.15 - 0.76)^2 + 57

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-57/5.7121 =a

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