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2 years ago

What transformation to the graph occurs when f(x) = x^2 is changed to f(x) = 3(x-2)^2?

Mathematics

1 answer:

miskamm [114]2 years ago

7 0

Answer:

Step-by-step explanation:

The -2 in the parentheses moves the graph horizontally 2 units to the right.

The 3 then stretches this graph vertically by a factor 3.

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What is the greatest common factor of 15 and 64

SCORPION-xisa [38]

15 = 3*5
64= 2*2*2*2*2*2

There are no common factors so the GCF is 1

5 0

2 years ago

Find the distance between (2, 2) and (-4, 4). Round answer to the nearest tenth.

Naya [18.7K]

Answer:

6.3

Step-by-step explanation:

First off, we need the distance formula, which is:

$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

If we plug in the points, we get:

$\sqrt{(-4 - 2)^2 + (4 - 2)^2}$

If we simplify everything under the square root, we get:

$\sqrt{(-6)^2 + (2)^2}$

$\sqrt{36 + 4}$

$\sqrt{40}$

In decimal, the answer is 6.3

6 0

3 years ago

The five-number summary of the ages of passengers on a cruise ship is listed below. Min 1 Q1 20Median 29 Q3 38Max 80 Consider th

Eddi Din [679]

Answer: b. Only statement (ii) is correct.

Step-by-step explanation:

The given five-number summary of the ages of passengers on a cruise ship is listed below.

Min 1 $Q_1$ 20 Median 29 $Q_3$ 38 Max 80

Inter-quartile range = $IQR=Q_3-Q_1=38-20=18$

According to the 1.5(IQR) criterion for outliers : An data value is an outlier if it lies below $Q_1 - 1.5(IQR)$ or above $Q_3 + 1.5(IQR)$ .

Here , $Q_1 - 1.5(IQR) =20-1.5(18)=-7$

$Q_3 + 1.5(IQR)=38+1.5(18)=65$

Since the minimum value> $Q_1 - 1.5(IQR)$ ( ∵ 1 > -7)

It means there is no value below $Q_1 - 1.5(IQR)$ , so there is no low -outlier.

⇒ Statement (i) "here is at least one passenger whose age is a low outlier. " is false.

But the maximum value > $Q_3 + 1.5(IQR)$ (∵ 85 > 65)

It means there are values above $Q_3 + 1.5(IQR)$ .

⇒Statement (ii) "There is at least one passenger whose age is a high outlier" is true.

Hence, the correct answer is b. Only statement (ii) is correct.

8 0

3 years ago

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .