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3 years ago

In which Quadrant are both cos and sin negative? he

Mathematics

1 answer:

Tomtit [17]3 years ago

8 0

Answer:

Third Quadrant

Step-by-step explanation:

1st Quadrant: All are positive

2nd Quadrant: Only sin is positive

3rd Quadrant: Only tan is positive

4th quadrant: Only cos is positive

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Find the intersection of the line and the circle given below y=-x-3 x^2+y^2=17

Salsk061 [2.6K]

Answer:

There are two points of intersection

(-4,1) and (1,-4)

Step-by-step explanation:

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3 years ago

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Find allgebraiclly the equation of the axis of symmetry and the vertex of the parabola represented by the equation y=-x2-2x+1

Slav-nsk [51]

To find the vertex use -b/2a (for the x value which is also the axis of symmetry)
So 2/-2 = -1 Therefore the axis of symmetry is x=-1 Plug in for the y value:
y= -(-1)^2 -2(-1) + 1 y=-1 + 2 + 1 y=2 Vertex (-1,2)

7 0

3 years ago

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Solving Rational Functions Hello I'm posting again because I really need help on this any help is appreciated!!

Greeley [361]

Answer:

x = √17 and x = -√17

Step-by-step explanation:

We have the equation:

$\frac{3}{x + 4} - \frac{1}{x + 3} = \frac{x + 9}{(x^2 + 7x + 12)}$

To solve this we need to remove the denominators.

Then we can first multiply both sides by (x + 4) to get:

$\frac{3*(x + 4)}{x + 4} - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}$

$3 - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}$

Now we can multiply both sides by (x + 3)

$3*(x + 3) - \frac{(x + 4)*(x+3)}{x + 3} = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

$3*(x + 3) - (x + 4) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

$(2*x + 5) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

Now we can multiply both sides by (x^2 + 7*x + 12)

$(2*x + 5)*(x^2 + 7x + 12) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}*(x^2 + 7x + 12)$

$(2*x + 5)*(x^2 + 7x + 12) = (x + 9)*(x + 4)*(x+3)$

Now we need to solve this:

we will get

$2*x^3 + 19*x^2 + 59*x + 60 = (x^2 + 13*x + 3)*(x + 3)$

$2*x^3 + 19*x^2 + 59*x + 60 = x^3 + 16*x^2 + 42*x + 9$

Then we get:

$2*x^3 + 19*x^2 + 59*x + 60 - ( x^3 + 16*x^2 + 42*x + 9) = 0$

$x^3 + 3x^2 + 17*x + 51 = 0$

So now we only need to solve this.

We can see that the constant is 51.

Then one root will be a factor of 51.

The factors of -51 are:

-3 and -17

Let's try -3

p( -3) = (-3)^3 + 3*(-3)^2 + +17*(-3) + 51 = 0

Then x = -3 is one solution of the equation.

But if we look at the original equation, x = -3 will lead to a zero in one denominator, then this solution can be ignored.

This means that we can take a factor (x + 3) out, so we can rewrite our equation as:

$x^3 + 3x^2 + 17*x + 51 = (x + 3)*(x^2 + 17) = 0$

The other two solutions are when the other term is equal to zero.

Then the other two solutions are given by:

x = ±√17

And neither of these have problems in the denominators, so we can conclude that the solutions are:

x = √17 and x = -√17

6 0

3 years ago

What is the perimeter of a quadrilateral with vertices at (3,-6),(8,-6),(3,-4), and (8,-4)

Eduardwww [97]

Answer:

im sorry i dont know but this is for a challenge

Step-by-step explanation:

3 0

2 years ago

A unit of electricity costs 13.2 pence. On average, tanya uses 90 units of electricity per week. She pays for her bill in 12 wee

Scrat [10]

Her electricity bill will be of 14256 pence.

Explanation

On average, Tanya uses 90 units of electricity per week.

So, the total units of electricity in 12 weeks will be: $(90*12)=1080$ units

One unit of electricity costs 13.2 pence.

Thus, the total cost for 1080 units will be: $(13.2*1080)pence= 14256$ pence.

So, her electricity bill will be of 14256 pence.

7 0

3 years ago

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