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Digiron [165]
2 years ago
4

Given f(x) and g(x) = f(k⋅x), use the graph to determine the value of k. Two lines labeled f of x and g of x. Line f of x passes

through points negative 4, 0 and 0, 4. Line g of x passes through points negative 2, 0 and 0, 4. −2 negative one half one half 2
Mathematics
1 answer:
jonny [76]2 years ago
4 0

Answer:

k=2

Step-by-step explanation:

the equation of  straight line passing through the two points (a , b) and (c , d) is  

Now f(x) passes through (-4 , 0) and (0 , 4)

the equation is  

y=x+4

Now g(x) passes through (-2 , 0) and (0 , 4)

the equation is  

y=2x+4

here f(x)=x+4 and g(x)=2x+4

clearly g(x)=f(2x)

therefore k=2

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What is the maximum number of solutions each of the following systems could have?
Marrrta [24]

Answer:

Two distinct concentric circles: 0 max solutions

Two distinct parabolas: 4 max solutions

A line and a circle: 2 max solution

A parabola and a circle: 4 max solutions

Step-by-step explanation:

<u>Two distinct concentric circles:</u>

The maximum number of solutions (intersections points) 2 distinct circles can have is 0. You can see an example of it in the first picture attached. The two circles are shown side to side for clarity, but when they will be concentric, they will have same center and they will be superimposed. So there can be ZERO max solutions for that.

<u>Two distinct parabolas:</u>

The maximum solutions (intersection points) 2 distinct parabolas can have is 4. This is shown in the second picture attached. <em>This occurs when two parabolas and in perpendicular orientation to each other. </em>

<u>A line and a circle:</u>

The maximum solutions (intersection points) a line and a circle can have is 2. See an example in the third picture attached.

<u>A parabola and a circle:</u>

The maximum solutions (intersection points) a parabola and a circle can have is 4. If the parabola is <em>compressed enough than the diameter of the circle</em>, there can be max 4 intersection points. See the fourth picture attached as an example.

4 0
2 years ago
Read 2 more answers
In ΔTUV, t = 6.6 inches, ∠U=71° and ∠V=36°. Find the length of u, to the nearest 10th of an inch.
natali 33 [55]

Answer:

hi

Step-by-step explanation:

5 0
2 years ago
Item 16 A rectangular garden is 6 feet long and 4 feet wide. A second rectangular garden has dimensions that are double the dime
____ [38]
P=2(L+W)
first one
P=2(6+4)
P=2(10)=20

doubled

12 and 8
P=2(12+8)
P=2(20)
P=40

the change is i20
percent change=change/original
change/original=20/20=1=100%


100% change
6 0
2 years ago
Simpsons, Inc. can sell 500 pairs of boots per week if they charge $100 Sales drop to 480 pairs per week if they charge $108. As
zhuklara [117]

Answer:

  a) (demand, price) = (460, 116), (520, 92)

  b) p = 300 -0.4q

  c) $300; 750 pairs

  d) 60 pairs; $28

Step-by-step explanation:

a) For each change in demand by 20 pairs, the corresponding price change is $8 in the opposite direction. 460 pairs is 20 <em>less</em> than 480 pairs, so the corresponding price is $8 <em>more</em> than the $108 price for 480 pairs. That is, (q, p) = (460, $116).

Similarly, 520 is 20 <em>more</em> pairs than the 500 sold for $100, so the corresponding price is $8 <em>less</em>. That is, (q, p) = (520, $92).

__

b) The two-point form of the equation of a line will do.

  p = (p2 -p1)/(q2 -q1)(q -q1) +p1

  p = (100 -108)/(500 -480)(q -480) +108 . . . . substitute given table values

  p = -8/20(q -480) +108

  p = -0.4q +300 . . . . . . . simplify

__

c1) When q=0, p = -0.4·0 +300 = 300

  A price of $300 will bring demand to zero.

c2) When p=0, q can be found from 0 = -0.4q +300.

  q = 300/0.4 = 750

  When boots are free, 750 pairs will be demanded.

__

d1) The derivative of price with respect to quantity is the coefficient of q in the equation:

  dp/dq = -0.4

Then ...

  dq/dp = 1/(-0.4) = -2.5

For a change in price of -$24, the change in quantity sold will be ...

  Δq = (dq/dp)·Δp = (-2.5)(-24) = 60

  60 more pairs will be sold if the price is reduced by $24.

d2) As for the previous question, ...

  Δp = (dp/dq)·Δq = (-0.4)(-70) = 28

  Raising the price by $28 will result in 70 fewer pairs sold.

__

Using derivatives is a fancy way to say the relationship between change in price and change in quantity is ...

  Δq : Δp = 20 : -8 (given in the problem)

  = 60 : -24 (part d1 -- multiply by 3)

  = -70 : 28 (part d2 -- multiply by -3.5)

3 0
2 years ago
In January, 2005 Anthony was preparing to purchase 100 shares of ABC stock. He was studying the literature and found four differ
Serga [27]

Answer:

The chart which is most relevant representation for Anthony's purposes is;

b. A graph has year on the x-axis, from 2001 to 2005, and value (dollars per share) on the y-axis, from 80 to 200 in increments of 40. A curve goes through (0, 170), (2003, 90), (2005, 130)

Step-by-step explanation:

From the question, we have;

The time at which Anthony was preparing to purchase shares = January, 2005

The number of shares Anthony was preparing to purchase = 100 shares

The information he found in the literature = Charts on ABC stock performance

The information required = The current price of shares of ABC stock

Therefore, the chart that is most relevant is a chart that gives the 2005 pricing  of ABC stock which is the graph that has a curve that goes through (0, 170), (2003, 90), (2005, 130)

7 0
2 years ago
Read 2 more answers
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