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svlad2 [7]
2 years ago
15

Which function in vertex form is equivalent to f(x) = x2 + 6x + 3?

Mathematics
2 answers:
Svetlanka [38]2 years ago
8 0
<span>f(x)=x^2+6x+3
     =(1/2.6)^2=3^2=9
f(x)=(x2^+6x+9)+3-9
     =(x+3)^2-6</span>
solong [7]2 years ago
7 0

Answer:

B. f(x) = (x + 3)2 − 6

Step-by-step explanation:

I just did this for "completing the square". Hope this helped!

You might be interested in
The 3rd degree Taylor polynomial for cos(x) centered at a = π 2 is given by, cos(x) = − (x − π/2) + 1/6 (x − π/2)3 + R3(x). Usin
Otrada [13]

Answer:

The cosine of 86º is approximately 0.06976.

Step-by-step explanation:

The third degree Taylor polynomial for the cosine function centered at a = \frac{\pi}{2} is:

\cos x \approx -\left(x-\frac{\pi}{2} \right)+\frac{1}{6}\cdot \left(x-\frac{\pi}{2} \right)^{3}

The value of 86º in radians is:

86^{\circ} = \frac{86^{\circ}}{180^{\circ}}\times \pi

86^{\circ} = \frac{43}{90}\pi\,rad

Then, the cosine of 86º is:

\cos 86^{\circ} \approx -\left(\frac{43}{90}\pi-\frac{\pi}{2}\right)+\frac{1}{6}\cdot \left(\frac{43}{90}\pi-\frac{\pi}{2}\right)^{3}

\cos 86^{\circ} \approx 0.06976

The cosine of 86º is approximately 0.06976.

8 0
2 years ago
last night, julie's pet hamster zoe kept julie awake for at least an hour running on her exercise wheel and scratching at the co
borishaifa [10]

Answer: x+y\geq 1, x>\frac{1}{4} ,  x\geq 2y and y\geq 0

Step-by-step explanation:

Here,  x represents the number of hours Zoe spent running on her wheel and y represents the number of hours spent scratching her cage.

Julie was awoke for at least an hour running on her exercise wheel and scratching the of her cage.

⇒ x+y\geq 1

She ran on her wheel at least twice as long as she scratched at the corners of her cage.

⇒ x\geq 2y

Also, She spent more than 1/4 hour running on her wheel.

⇒ x>\frac{1}{4}

And, we know that number of hours can not be negative.

⇒  y\geq 0

Therefore, the complete system of inequality which shows the given situation is,

x+y\geq 1, x>\frac{1}{4} and x\geq 2y, y\geq 0

Note: the feasible region ( covered by the given system) is shown in the below graph.


6 0
2 years ago
Find the distance from (4, −7, 6) to each of the following.
LenKa [72]

Answer:

(a) 6 units

(b) 4 units

(c) 7 units

(d) 9.22 units

(e) 7.21 units

(f) 8.06 units

Step-by-step explanation:

The distance d from one point (x₁, y₁, z₁) to another point (x₂, y₂, z₂) is given by;

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Now from the question;

<em>(a) The distance from (4, -7, 6) to the xy-plane</em>

The xy-plane is the point where z is 0. i.e

xy-plane = (4, -7, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, -7, 0)</em>

d = √[(4 - 4)² + (-7 - (-7))² + (0 - 6)²]

d = √[(0)² + (0)² + (-6)²]

d = √(-6)²

d = √36

d = 6

Hence, the distance to the xy plane is 6 units

<em>(b) The distance from (4, -7, 6) to the yz-plane</em>

The yz-plane is the point where x is 0. i.e

yz-plane = (0, -7, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 6)</em>

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 6)²]

d = √[(4)² + (0)² + (0)²]

d = √(4)²

d = √16

d = 4

Hence, the distance to the yz plane is 4 units

<em>(c) The distance from (4, -7, 6) to the xz-plane</em>

The xz-plane is the point where y is 0. i.e

xz-plane = (4, 0, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 6)</em>

d = √[(4 - 4)² + (-7 - 0)² + (6 - 6)²]

d = √[(0)² + (-7)² + (0)²]

d = √[(-7)²]

d = √49

d = 7

Hence, the distance to the xz plane is 7 units

<em>(d) The distance from (4, -7, 6) to the x axis</em>

The x axis is the point where y and z are 0. i.e

x-axis = (4, 0, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 0)</em>

d = √[(4 - 4)² + (-7 - 0)² + (6 - 0)²]

d = √[(0)² + (-7)² + (6)²]

d = √[(-7)² + (6)²]

d = √[(49 + 36)]

d = √(85)

d = 9.22

Hence, the distance to the x axis is 9.22 units

<em>(e) The distance from (4, -7, 6) to the y axis</em>

The x axis is the point where x and z are 0. i.e

y-axis = (0, -7, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 0)</em>

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 0)²]

d = √[(4)² + (0)² + (6)²]

d = √[(4)² + (6)²]

d = √[(16 + 36)]

d = √(52)

d = 7.22

Hence, the distance to the y axis is 7.21 units

<em>(f) The distance from (4, -7, 6) to the z axis</em>

The z axis is the point where x and y are 0. i.e

z-axis = (0, 0, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, 0 6)</em>

d = √[(4 - 0)² + (-7 - (0))² + (6 - 6)²]

d = √[(4)² + (-7)² + (0)²]

d = √[(4)² + (-7)²]

d = √[(16 + 49)]

d = √(65)

d = 8.06

Hence, the distance to the z axis is 8.06 units

5 0
2 years ago
Match the following items.
Pani-rosa [81]

This problem can be solved using probability, the equation of the probability of an event A is P(A)= favorable outcomes/possible outcomes. The interception of two probable events is P(A∩B)= P(A)P(B).

There are 12 black marbels, 10 red marbles, and 18 white marbels, all the same size. If two marbles are drawn from the jar without being replaced.

The total of the marbles is 40.

If two marbles are drawn from the jar without being replaced, what would the probability be:

1. of drawing two black marbles?

The probability of drawing one black marble is (12/40). Then, the probability of drawing another black marble after that is (11/39) due we drawing one marble before.

P(Black∩Black) = (12/40)(11/39) = 132/1560, simplifying the fraction:

P(Black∩Black) = 11/130

2. of drawing a white, then a black marble?

The probability of drawing one white marble is (18/40). Then, the probability of drawing then a black marble after that is (12/39) due we drawed one marble before.

P(White∩Black) = (18/40)(12/39) = 216/1560, simplifying the fraction:

P(White∩Black) = 9/65

3. of drawing two white marbles?

The probability of drawing one white marble is (18/40). Then, the probability of drawing another white marble after that is (17/39) due we drawed one marble before.

P(White∩White) = (18/40)(17/39) = 306/1560, simplifying the fraction:

P(White∩White) = 51/260

4. of drawing a black marble, then a red marble?

The probability of drawing one black marble is (12/40). Then, the probability of drawing then a red marble after that is (10/39) due we drawed one marble before.

P(Black∩Red) = (12/40)(10/39) = 120/1560, simplifying the fraction:

P(Black∩Red) = 1/13

4 0
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