$2x-y=10\Rightarrow 2x-10=y \\ 
\Longrightarrow 3x-2y=8\Rightarrow 3x-2(2x-10)=8 \\ 
3x-4x+20=8 \\ -x=8-20=-12 \\ x=12 \\ \\ y=2(12)-10=24-10 \\ y=14 \\ \\ (x,y)=(12,14)$

8 0

3 years ago

Approximate the square root of 77

GarryVolchara [31]

Answer:

abt 8.77

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Pls help extra points and mark brainlist

kipiarov [429]

Answer:

x-4=19

Step-by-step explanation:

Hope that helps!

5 0

3 years ago

Read 2 more answers

How can csc^2-cot^2 = 1?

Softa [21]

ANSWER

By simplifying the left hand side using the Pythagorean Identity.

EXPLANATION

The given identity is

$\csc^{2} (x) - \cot^{2} (x) = 1$

Take the left hand side and simplify to get the right hand side.

$\csc^{2} (x) - \cot^{2} (x) = \frac{1}{\sin^{2} (x)} - \frac{ \cos^{2} (x)}{\sin^{2} (x)}$

Collect LCM for the denominators.

$\csc^{2} (x) - \cot^{2} (x) = \frac{1 - \cos^{2} (x)}{\sin^{2} (x)}$

Recall the Pythagorean Identity.

$\cos^{2} (x) + \sin^{2} (x) = 1$

This implies that:

$1 - \cos^{2} (x) = \sin^{2} (x)$

We substitute this to get,

$\csc^{2} (x) - \cot^{2} (x) = \frac{\sin^{2} (x)}{\sin^{2} (x)}$

$\csc^{2} (x) - \cot^{2} (x) = 1$

6 0

3 years ago