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erastova [34]
1 year ago
5

f the pattern in the table is extended to represent more equivalent ratios for 2:6, which pair of numbers would be in the column

s? A multiplication table. In the column labeled 2, the numbers 2, 4, 6, 8, 10, 12, 14, 16, and 18 are highlighted. In the column labeled 6, the numbers 6, 12, 18, 24, 30, 36, 42, 48, and 54 are highlighted. 20 would be in the column for 2, and 60 would be in the column for 6. 20 would be in the column for 6, and 60 would be in the column for 2. 20 would be in the column for 2, and 56 would be in the column for 6. 20 would be in the column for 6, and 56 would be in the column for 2. please hurry im in a rush
Mathematics
1 answer:
olasank [31]1 year ago
3 0

The pair of numbers that would be in the columns, considering the proportional relationship, is given as follows:

20 would be in the column for 2, and 60 would be in the column for 6.

<h3>What is a proportional relationship?</h3>

A proportional relationship is a special linear function, with intercept having a value of zero, in which the output variable is obtained with the multiplication of the input variable and the constant of proportionality k, as shown as follows:

y = kx

The table is extended to represent more equivalent ratios for 2:6, hence the constant of the relationship is given as follows:

k = 6/2 = 3.

Hence the equation is:

y = 3x.

The values given by each column are given as follows:

  • Column 2: values of x.
  • Column 6: values of y.

When x = 20, the numeric value of the relationship is of:

y = 3 x 20 = 60.

Hence the first option is correct.

More can be learned about proportional relationships at brainly.com/question/10424180

#SPJ1

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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.
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Answer:

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = 7

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = -5\sqrt 2

Step-by-step explanation:

Given

N = (-7,-1) --- terminal side of \varnothing

Required

Determine the values of trigonometric functions of \varnothing.

For \varnothing, the trigonometry ratios are:

sin\ \varnothing = \frac{y}{r}       cos\ \varnothing = \frac{x}{r}       tan\ \varnothing = \frac{y}{x}

cot\ \varnothing = \frac{x}{y}       sec\ \varnothing = \frac{r}{x}       csc\ \varnothing = \frac{r}{y}

Where:

r^2 = x^2 + y^2

r = \sqrt{x^2 + y^2

In N = (-7,-1)

x = -7 and y = -1

So:

r = \sqrt{(-7)^2 + (-1)^2

r = \sqrt{50

r = \sqrt{25 * 2

r = \sqrt{25} * \sqrt 2

r = 5 * \sqrt 2

r = 5 \sqrt 2

<u>Solving the trigonometry functions</u>

sin\ \varnothing = \frac{y}{r}

sin\ \varnothing = \frac{-1}{5\sqrt 2}

Rationalize:

sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

sin\ \varnothing = \frac{-\sqrt 2}{5*2}

sin\ \varnothing = \frac{-\sqrt 2}{10}

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{x}{r}

cos\ \varnothing = \frac{-7}{5\sqrt 2}

Rationalize

cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}

cos\ \varnothing = \frac{-7\sqrt 2}{10}

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{y}{x}

tan\ \varnothing = \frac{-1}{-7}

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = \frac{x}{y}

cot\ \varnothing = \frac{-7}{-1}

cot\ \varnothing = 7

sec\ \varnothing = \frac{r}{x}

sec\ \varnothing = \frac{5\sqrt 2}{-7}

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = \frac{r}{y}

csc\ \varnothing = \frac{5\sqrt 2}{-1}

csc\ \varnothing = -5\sqrt 2

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