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2 years ago

Solve for y. 2x-3y=9

Mathematics

1 answer:

cluponka [151]2 years ago

6 0

Answer:

y = 2/3x - 3

Step-by-step explanation:

Step 1: Write equation

2x - 3y = 9

Step 2: Solve for y

Subtract 2x on both sides: -3y = 9 - 2x
Divide both sides by -3: y = -3 + 2/3x
Rewrite: y = 2/3x - 3

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How do I get (tan^2(x)-sin^2(x))/tan(x) equal to (sin^2(x))/cot(x)

shepuryov [24]

$LHS\\ \\ =\frac { \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } }{ \tan { x } } \\ \\ =\frac { 1 }{ \tan { x } } \left( \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } \right)$

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^{ 2 }{ x } }{ \cos ^{ 2 }{ x } } -\frac { \sin ^{ 2 }{ x\cos ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \right) \\ \\ =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^{ 2 }{ x-\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } } }{ \cos ^{ 2 }{ x } } \right)$

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\left( 1-\cos ^{ 2 }{ x } \right) } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\cdot \sin ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } \sin ^{ 4 }{ x } }{ \sin { x\cos ^{ 2 }{ x } } } \\ \\ =\frac { \sin ^{ 3 }{ x } }{ \cos { x } }$

$\\ \\ =\sin ^{ 2 }{ x } \cdot \frac { \sin { x } }{ \cos { x } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \frac { \cos { x } }{ \sin { x } } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \cot { x } } \\ \\ =\frac { \sin ^{ 2 }{ x } }{ \cot { x } } \\ \\ =RHS$

8 0

2 years ago

You purchased xyz common shares for $45 per share one year ago. one year later aftering receiving $4 dividend per share, you sol

Pepsi [2]

The return for the investment is C, or 15.56%. You add the dividend and the capital gain, $4+$3=$7, and then divide by the initial investment price, or $45. 7/45=.1556, or 15.56%.

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2 years ago

Check out if I'm right Plz explain..if u have a different answer

Ronch [10]

Answer:

A C

Step-by-step explanation:

A is true.The numerator does have 3 terms. The constant term (the 3 at the end) is still a term and counts as a term.

A constant term does not count as a coefficient so b is not right, just as you have indicated by not underlining it.

C is true. In fact, 2 is the leading coefficent of the denominator.

Either A is true or D is, but I don't think they both are. You need a modifier to claim that D is true. If the statement said the numerator has at least 2 terms then both A and D would be true. Without the at least, you have to pick one and the one I choose is A

This is the tough one. Very sticky. You would think it is true, but it isn't. A constant term is not a coefficient. Coefficients only count when they are with "x"s.

8 0

2 years ago

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Evaluate the function f(x)=x^2+6x+2 at the given values of the independent variable and simplify.

____ [38]

Answer:

a) 18

b)x^2+10x+18

c)x^2 -6x+2

Step-by-step explanation:

This is a case of plugging in the value into f(x).

a) f(-8)= -8^2 + 6(-8) +2

f(-8)= 64 + (-48) +2

f(-8)=64 + (-46)

f(-8)=18

b) f(x+2)= (x+2)^2+6(x+2)+2

So here I'll take a break to explain what's going on, because x+2 is a binomial meaning two terms and it is being squared I have to multiply the whole thing by itself. Meaning: (x+2) x (x+2), this is also known as foiling!! and for the next part its distributing 6 into x and 2.

f(x+2)= x^2+4x+4+6x+12+2

I'll reorder it

f(x+2)= x^2+4x+6x+12+2+4

f(x+2)= x^2+10x+18

c) f(-x)= -x^2+6(-x) +2

f(-x)= x^2 -6x+2

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2 years ago

Is -3/5 greater than, less than, or equal to -0.6

Vlad1618 [11]

The answer is equal to -0.6

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2 years ago

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