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ANTONII [103]
3 years ago
12

Which function has a minimum and is transformed to the right and down from the parent function, f(x) = x2? g(x) = –9(x + 1)2 – 7

g(x) = 4(x – 3)2 + 1 g(x) = –3(x – 4)2 – 6 g(x) = 8(x – 3)2 – 5
Mathematics
2 answers:
zalisa [80]3 years ago
6 0

Answer:

g(x) = 8\cdot (x-3)^{2}-5

Step-by-step explanation:

Given that parent function represents a parabola, the standard form with a vertex at (h,k) is now described:

y-k = C\cdot (x-h)^{2}

y = C \cdot (x-h)^{2} + k

Where:

x, y - Independent and dependent variables, dimensionless.

h, k - Horizontal and vertical component of the vertex, dimensionless.

C - Vertex factor, dimensionless. (If C > 0, then vertex is an absolute minimum, but if C < 0, there is an absolute maximum).

After reading the statement of the problem, the following conclusion are found:

1) New function must have an absolute minimum: C > 0

2) Transformation to the right: h > 0.

3) Transformation downwards: k < 0

Hence, the right choice must be g(x) = 8\cdot (x-3)^{2}-5.

Lorico [155]3 years ago
4 0

Answer:

Choice D

Step-by-step explanation:

Took the test

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A 140 kg camera is suspended by two wires over a 40 metre wide football field to get shots of the action from above. At one poin
Katen [24]

Answer:

Please red the answer below

Step-by-step explanation:

In order to determine the length of each cable you use the Newton second law for each component of the forces involved in the situation.

For the x component you have:

T_1cos\theta_1-T_2cos\theta_2=0           (1)

T1: tension of the first cable = 1500N

T2: tension of the second cable = 800N

θ1: angle between the horizontal and the first cable

θ2: angle between the horizontal and the first cable

For the y component you have:

T_1sin\theta_1+T_2sin\theta_2-W=0               (2)

W: weight of the camera = Mg = (140kg)(9.8m/s^2) = 1372N

You can squared both equations (1) and (2) and the sum the two equations:

T_1^2cos^2\theta_1=T_2^2cos^2\theta_2\\\\T_1^2sin^2\theta_1=T_2^2sin^2\theta_2-2WT_2sin\theta_2+W^2

Then, you sum the equations:

T_1^2(cos^2\theta_1+sin^2\theta_1)=T_2^2(sin^2\theta_2+cos^2\theta_2)-2Wsin\theta_2+W^2        (3)

Next, you use the following identity:

sin^2\theta+cos^2\theta=1

and you obtain in the equation (3):

T_1^2=T_2^2-2WT_2sin\theta_2+W^2\\\\sin\theta_2=\frac{T_2^2-T_1^2+W^2}{2WT_2}=\frac{(800N)^2-(1500)^2+(1372N)^2}{2(800N)(1372N)}=0.066\\\\\theta_2=sin^{-1}0.066=27.23\°

With this values you can calculate the value of the another angle, by using the equation (1):

\theta_1=cos^{-1}(\frac{T_2cos(27.23\°)}{T_1})=cos^{-1}(\frac{(800N)(cos27.23\°)}{1500N})\\\\\theta_1=61.69\°

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l_1cos\theta_1=40-d\\\\l_2cos\theta_2=d\\\\

d: distance to the right side of the field

By using the cosine law you can fins a system of equation and then you can calculate the values of l1 and l2.

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