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3 years ago

Is this right?

Mathematics

1 answer:

Mumz [18]3 years ago

4 0

Answer:

Q1.

AC is the greatest side, therefore is hypotenuse. AB and BC are legs.

B is 90 deg as the two legs cross at this point.

<u>Ratios:</u>

sin = opposite / hypotenuse
cos = adjacent / hypotenuse
tan = opposite / adjacent

BC is opposite to A and adjacent to C

AB is adjacent to A and opposite to C

sin A = cos C = 8/10 = 4/5
cos A = sin C = 6/10 = 3/5
tan A = cot C = 8/6 = 4/3
tan C = cot A = 6/8 = 3/4

Q2.

a) Answered above

b) Answered above

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Complete the square to rewrite x^2+y^2+2x+6y-6=0 in graphing form

Ksivusya [100]

Answer:

(x + 1)² + (y + 3)² = 16

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

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Given

x² + y² + 2x + 6y - 6 = 0

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x² + 2x + y² + 6y = 6

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3 years ago

What is a solution to the equation 3 / m + 3 - M / 3 - M equals m^2 + 9 / m^2-9?

Mnenie [13.5K]

Answer: Last option.

Step-by-step explanation:

Given the equation:

$\frac{3}{m+3}-\frac{m}{3-m}=\frac{m^2+9}{m^2-9}$

Follow these steps to solve it:

- Subtract the fractions on the left side of the equation:

$\frac{3(3-m)-m(m+3)}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{9-3m-m^2-3m}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}$

- Using the Difference of squares formula ( $a^2-b^2=(a+b)(a-b)$ ) we can simplify the denominator of the right side of the equation:

$\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{(m+3)(m-3)}$

- Multiply both sides of the equation by $(m+3)(3-m)$ and simplify:

$\frac{(-m^2-6m+9)(m+3)(3-m)}{(m+3)(3-m)}=\frac{(m^2+9)(m+3)(3-m)}{(m+3)(m-3)}\\\\-m^2-6m+9=\frac{(m^2+9)(3-m)}{(m-3)}$

- Multiply both sides by $m-3$ :

$(-m^2-6m+9)(m-3)=\frac{(m^2+9)(3-m)(m-3)}{(m-3)}\\\\(-m^2-6m+9)(m-3)=(m^2+9)(3-m)$

- Apply Distributive property and simplify:

$(-m^2-6m+9)(m-3)=(m^2+9)(3-m)\\\\-m^3-6m^2+9m+3m^2+18m-27=3m^2+27-m^3-9m\\\\-m^3-3m^2+27m-27+m^3-3m^2+9m-27=0\\\\-6m^2+36m-54=0$

- Divide both sides of the equation by -6:

$\frac{-6m^2+36m-54}{-6}=\frac{0}{-6}\\\\m^2-6m+9=0$

- Factor the equation and solve for "m":

$(m-3)^2=0\\\\m=3$

In order to verify it, you must substitute $m=3$ into the equation and solve it:

$\frac{3}{3+3}-\frac{3}{3-3}=\frac{3^2+9}{3^2-9}\\\\\frac{3}{6}-\frac{3}{0}=\frac{18}{0}$

<em>NO SOLUTION</em>

7 0

2 years ago

Whats the quotient of simplest from for 5 wholes divide by 2\9 ?

posledela

Keep Change Flip!!!
Keep: the first fraction 5 or 5/1
Change: Division sign to multiplication sign
Flip (reciprocal): the second fraction
So you get 5/1*9/2= 45/2 which simplified to 22 1/2

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3 years ago