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lorasvet [3.4K]

2 years ago

8

Simplify 7-5x-(3+5x)

1 answer:

galben [10]2 years ago

8 0

Answer:

-10x +4

Step-by-step explanation:

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What is the answer to 12-2u=9u+45

Mars2501 [29]

Answer:

u = -3

Step-by-step explanation:

12 - 2u = 9u + 45

2u - 12 = -9u - 45

2u + 9u = 12 - 45

11u = -33

u = -33/11

u = -3

5 0

3 years ago

3(x-8) = -29.7 please solve for x

GrogVix [38]

Answer:

The answer is x= -1.9.

Use the 3 to multiply the number and the letter in the bracket... which is...

3x-24=-29.7.

3x=-29.7+24.

3x=-5.7.

x=-5.7/3.

x=-1.9.

6 0

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:

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

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

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

$cot\ \varnothing = 7$

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

$csc\ \varnothing = -5\sqrt 2$

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

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

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

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

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

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

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

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

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

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

$cot\ \varnothing = 7$

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

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

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

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

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

$csc\ \varnothing = -5\sqrt 2$

3 0

2 years ago

I need help solving 2x+7=x+3

mart [117]

2x+7=x+3

2x+7=3
-×

x+7=3
-7 -7

x=-4

8 0

3 years ago

Read 2 more answers

Find the leg of each isosceles right triangle when the hypotenuse is 5 square root 6

kozerog [31]

Hypotenuse = 5√6

Let the length of each leg be x, since it is Isosceles .

x² + x² = (5√6)²

2x² = (5 * 5 * √6 * √6)

2x² = 150

x² = 150/2

x² = 75

x = √75

x = √(25 * 3)

x = √25 * √3

x = 5√3

5 0

3 years ago