Identify the center and radius of each equation. (x

(-6x+3)+(5x-4)<br><br> If you in the 7th grade or up you can answer this question

Otrada [13]

Answer:

-1x + -1

Step-by-step explanation:

(-6x+3)+(5x-4)

-1x + -1

7 0

2 years ago

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

7 0

2 years ago

Which expresses 2.7 billion in scientific notation

liraira [26]

Answer: A 2.7 x 10^9

Explanation: brain go brrrrr

8 0

2 years ago

Read 2 more answers

What are the solutions of the equation x^4-9x^2+8=0? Use u substitution to solve.

vodka [1.7K]

Answer:

The answer is the second option

Step-by-step explanation:

Let u=x^2.

$( {x}^{2} ) {}^{2} - 9( {x}^{2} ) + 8( {x}^{2} ) {}^{0}$

${u}^{2} - 9u + 8$

$(u - 1)(u - 8)$

$u = 1$

$u = 8$

Set this equal to x^2

${x}^{2} = 1$

$x = 1$

or

$x = - 1$

${x}^{2} = 8$

$x = 2 \sqrt{2}$

or

$x = - 2 \sqrt{2}$

4 0

1 year ago

Need help what the answer

stealth61 [152]

I think that it might be 62,100 because the 6 is in the ten thousands place, and if you divide it by 10, 60,000/10, then it would be 6000. 60,000 is 10 times more than 6,000. Correct me if I'm wrong.

8 0

3 years ago

Identify the center and radius of each equation. (x - 1)² + y² +81