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3 years ago

Solve -9(t - 2) = 4(1-15).

Mathematics

2 answers:

storchak [24]3 years ago

8 0

Answer:

t = 74 / 9 ~= 8.222

Step-by-step explanation:

-9 ( t - 2 ) = 4 ( 1 - 15 )

-9 * t + -9 * -2 = 4 * 1 + 4 * -15

-9t + 18 = 4 + -60

-9t + 18 = -56

-9t = -56 + -18

-9t = -74

t = 74 / 9

Alternatively:

-9 ( t - 2 ) = 4 ( 1 - 15 )

-9 * t + -9 * -2 = 4 ( - 14 )

-9t + 18 = -56

-9t = -56 + -18

-9t = -74

t = 74 /9

Cheers.

snow_lady [41]3 years ago

5 0

Answer:

t = 74 / 9

Step-by-step explanation:

-9(t - 2) = 4(1 - 15)

-9t + 18 = 4 - 60

-9t + 18 = -56

-18 -18

-9t = -74

/-9 /-9

t = 74 / 9

t = 8.22

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2 years ago

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 ∅.

Natali [406]

Answer:

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

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

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

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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:

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

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

Solving the trigonometry functions

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

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

Rationalize:

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

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

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Rationalize

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

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

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

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

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

$cot\ \varnothing = 7$

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

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

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

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

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

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