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2 years ago

Given that the point (1,2) lies on y = f(x), find the corresponding point on the image function of y = 3f(2x).

Mathematics

1 answer:

lana66690 [7]2 years ago

7 0

The point corresponding to (1, 2) on the graph of the function 3f(2x) is (3,6 ).

<h3>What is a function?</h3>

Function is a type of relation, or rule, that maps one input to specific single output.

Since, we are given a function f(x) such that the point (1,2) lie on the graph of the function f(x).

Now, we are given a new function g(x) as 3f(2x) i.e. g(x) = 3f(2x).

Clearly, this new function g(x) is a dilation of f(x) by a scale factor of '3'.

Hence, every coordinates of the this new function will get multiplied by the scale factor.

i.e. (3×1, 3× 2)=(3, 6)

Hence, the point corresponding to the point (1,2) on the graph of f(x) will convert to the point (3, 6) on the graph of g(x)= 3 f(2x).

Learn more about function here:

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Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, $\mu_X$ represent the population mean for Brand B and let $\mu_Y$ represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $Xbar$ = Sample mean for Brand B data = 36.9

$Ybar$ = Sample mean for Brand A data = 32.9

$n_1$ = Sample size for Brand B data = 13

$n_2$ = Sample size for Brand A data = 9

$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ , (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ]

[ $(36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ , $(36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

4 0

2 years ago

Stephanie has $4,500 in her saving account.

AfilCa [17]

so whats the question

7 0

3 years ago

Darrin surveyed a random sample of 10 students from his science class about their favorite types of TV shows. Five students like

Pepsi [2]

In my opinion, Darrin's inference is wrong because according to given question, "Darrin surveyed a random sample of 10 students from his science class about their favorite types of TV shows."

This line provides the information that the survey is taken randomly. Also, if Darrin had taken some other students, then the ineference of other new students compared with previously surveyed students will be different.

This frankly tells that the probability is different always.

Therefore, Darrin's inference is wrong or invalid.

4 0

2 years ago

The graph below shows the function f(x)=x-3/x^2-2x-3 which statement is true

Ugo [173]

Answer:

The correct option is A.

Step-by-step explanation:

Domain:

The expression in the denominator is x^2-2x-3

x² - 2x-3 ≠0

-3 = +1 -4

(x²-2x+1)-4 ≠0

(x²-2x+1)=(x-1)²

(x-1)² - (2)² ≠0

∴a²-b² =(a-b)(a+b)

(x-1-2)(x-1+2) ≠0

(x-3)(x+1) ≠0

x≠3 for all x≠ -1

So there is a hole at x=3 and an asymptote at x= -1, so Option B is wrong

Asymptote:

x-3/x^2-2x-3

We know that denominator is equal to (x-3)(x+1)

x-3/(x-3)(x+1)

x-3 will be cancelled out by x-3

1/x+1

We have asymptote at x=-1 and hole at x=3, therefore the correct option is A....

4 0

3 years ago

Read 2 more answers

Elizabith uses 23.25 ounces of granola and 10.5 ounces of raisins for 15 servings of tril mix if each serving contains the same

d1i1m1o1n [39]

1.55 ounces of granola and 0.7 ounces of raisins in each serving

5 0

2 years ago