Evaluate: 542 × 31

faltersainse [42]

Can u show a picture 0.0

6 0

3 years ago

suppose f(x)=3x+5. describe how the graph of each function compares to f. 1) g(x)=f(x)+12 2) h(x)=f(x)-7 3) g(x)=f(x=8) 4) h(x)=

Lubov Fominskaja [6]

1)

g(x) = f(x)+12 represents translating f(x) 12 units up to get g(x). This is because y = f(x), so we're effectively adding 12 to each y coordinate of the points on f(x) to have them move to the points on g(x).

note: the term "translating" is another way of saying "shifting" or "sliding".

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2)

h(x) = f(x)-7 means translating f(x) 7 units down to get h(x). We use the same reasoning as in problem 1. The general form is h(x) = f(x)+k, and in this case, k = -7. The negative k value tells us to shift down.

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3)

g(x) = f(x+8) means we translate 8 units to the left. The general template is g(x) = f(x-h). If h > 0, then we shift right. If h < 0, then we shift left. The amount that is shifted is equal to the absolute value of h.

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4)

h(x) = f(x-14) tells us to shift f(x) 14 units to the right. Same idea as problem 3, but now h = 14.

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5)

g(x) = 4*f(x) means we vertically stretch f(x) by a factor of 4. Since y = f(x), we are effectively doing g(x) = 4*f(x) = 4*y, which pulls the y coordinates upward by a factor of 4. A point like (1,2) will jump to (1,8) after multiplying the y coordinate by 4.

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6)

g(x) = f(5x) tells us to horizontally compress f(x) by a factor of 5. The general template is g(x) = f(a*x) where the 'a' determines horizontal stretching or compression. If $0 \le a \le 1$ , then we have horizontal stretching going on. If $a > 1$ , then we'll have horizontal compression.

6 0

3 years ago

Which of the following operations is true regarding relative frequency distributions? Multiple choice question. No two classes c

pshichka [43]

Answer:

The relative frequency is found by dividing the class frequencies by the total number of observations

Step-by-step explanation:

Relative frequency measures how often a value appears relative to the sum of the total values.

An example of how relative frequency is calculated

Here are the scores and frequency of students in a maths test

Scores (classes) Frequency Relative frequency

0 - 20 10 10 / 50 = 0.2

21 - 40 15 15 / 50 = 0.3

41 - 60 10 10 / 50 = 0.2

61 - 80 5 5 / 50 = 0.1

81 - 100 <u> 10</u> 10 / 50 = <u>0.2</u>

50 1

From the above example, it can be seen that :

two or more classes can have the same relative frequency
The relative frequency is found by dividing the class frequencies by the total number of observations.
The sum of the relative frequencies must be equal to one
The sum of the frequencies and not the relative frequencies is equal to the number of observations.

4 0

3 years ago

Abigail tosses a coin off a bridge into the stream below. The distance, in feet, the coin above the water is modeled by the equa

melomori [17]

Answer:

7 seconds

Step-by-step explanation:

Distance of the coin above the water is modeled by the equation

$y=-16x^2+96x+112$

where $x$ is time taken to cover the distance

Now equating with zero.

$0=-16x^2+96x+112\\\Rightarrow 16x^2-96x-112=0\\\Rightarrow x=\dfrac{-\left(-96\right)\pm \sqrt{\left(-96\right)^2-4\times \:16\left(-112\right)}}{2\times \:16}\\\Rightarrow x=7, -1$

So, time taken by the coin to hit the water is 7 seconds.

6 0

3 years ago

A local grocery store received 6 5/6 crates of strawberries. the produce workers had to throw away 1 7/8 crates of strawberries

BigorU [14]

Answer:

D. 4 23/24

Step-by-step explanation:

1 7/8 = 1 21/24

6 5/6 = 6 20/24

6 - 1 = 5

20 - 21 = -1

5 - 1/24 = 4 23/24

3 0

3 years ago