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3 years ago

What is the value of x in the following equation?

Mathematics

2 answers:

mihalych1998 [28]3 years ago

6 0

Answer:

-2

Step-by-step explanation:

-3x - 2x = 8 + 2

-5x = 10

x = -2

docker41 [41]3 years ago

4 0

Answer:

$x=-2$

Step-by-step explanation:

To cancel out "- 2", we add 2 into both sides.

$-3x(-2+2)=2x+(8+2)$

$-3x=2x+10$

Now, we need to get rid of 2x so we can have one x.

Let's subtract 2x from both sides.

$(-3x-2x)=(2x-2x)+10$

$-5x=10$

To get x by itself, we will divide both sides by -5.

$\frac{-5x}{-5} =\frac{10}{-5}$

$x=-2$

The value of x in the given equation is -2. I do not see a -2 in the answer choices.

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In which equation is x the independent variable? A. x=7y B. y=7x C. y+7=x D. y/7=x

Arturiano [62]

Answer:

B is the answer.

A) No because the value of x depends on what 7•y is

C) No because x depends on the value of y+7

D) No because the x depends on the value of y/7

In other words, on B, x doesn't depend on anything, your just given the value of it. Don't know how else to explain it, hope you get it.

7 0

3 years ago

Read 2 more answers

Find the exact values of the six trigonometric functions of the given angle. If any are not defined, say "not defined." Do not u

Solnce55 [7]

Answer:

Step-by-step explanation:

Let's take a look at the given angle 135°

The sketch of the angle which corresponds to $-\dfrac{3\pi}{4}$ unit circle and can be seen in the attached image below;

The trigonometric ratios are as follows for an angle θ on the unit circle:

Trigonometric ratio related ratio on coordinate axes

sin θ $\dfrac{y}{1}$

cos θ $\dfrac{x}{1}$

tan θ $\dfrac{y}{x}$

csc θ $\dfrac{1}{y}$

sec θ $\dfrac{1}{x}$

cot θ $\dfrac{x}{y}$

From the sketch of the image attached below;

The six trigonometric ratio for 135° can be expressed as follows:

$sin (-\dfrac{3\pi}{4})= \dfrac{y}{1}$

$sin (-\dfrac{3\pi}{4})=- \dfrac{\sqrt{2}}{2}$

$cos (-\dfrac{3\pi}{4})= \dfrac{x}{1}$

$cos (-\dfrac{3\pi}{4})= -\dfrac{\sqrt{2}}{2}$

$tan (-\dfrac{3\pi}{4})= \dfrac{y}{x}$

$tan (-\dfrac{3\pi}{4})= \dfrac{-\dfrac{\sqrt{2}}{2}}{-\dfrac{\sqrt{2}}{2}}$

$tan (-\dfrac{3\pi}{4})= -\dfrac{\sqrt{2}}{2}} \times {-\dfrac{2}{\sqrt{2}}$

$tan (-\dfrac{3\pi}{4})= 1$

$csc (-\dfrac{3\pi}{4})= \dfrac{1}{y} \\ \\ csc (-\dfrac{3\pi}{4})=\dfrac{1}{-\dfrac{\sqrt{2}}{2}} \\ \\ csc=1 \times -\dfrac{2}{\sqrt{2}} \\ \\csc =-\sqrt{2}$

$sec (-\dfrac{3 \pi}{4})=\dfrac{1}{x} \\ \\ sec = \dfrac{1}{(-\dfrac{\sqrt{2}}{2})} \\ \\ sec = 1 \times -\dfrac{2}{\sqrt{2}} \\ \\ sec = - \sqrt{2}$

$cot(-\dfrac{3 \pi}{4}) = \dfrac{x}{y} \\ \\ cot(-\dfrac{3 \pi}{4}) = \dfrac{-\dfrac{\sqrt{2}}{2} }{-\dfrac{\sqrt{2}}{2}} \\ \\ cot(-\dfrac{3 \pi}{4})= -\dfrac{\sqrt{2}}{2} } \times {-\dfrac{2}{\sqrt{2}}} \\ \\ cot (-\dfrac{3 \pi}{4}) = 1$