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3 years ago

Pleeaaassseeeees help,

Mathematics

1 answer:

sattari [20]3 years ago

5 0

Hey there!

Since you probably have better things to do, your answer would be A: π(3)²

To back up my answer, the formula to find the area of a circle is π · R² where R = Radius. In that formula then, you just plug everything in. Since the radius is 3, you would put 3 into the "( )". Finally, just plug the rest of the equation in!

I'm open to any question or comment!

God Bless!

-X8lue83rryX

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PLEASE HELP GUYS i am struggling so much, two questions

ankoles [38]

Answer: the equation of the standard parabola

1) $(y-6)^2 = 4 (x-1)$

The equation of the standard parabola

2) (x+5)^2 = 16(y-2)

Step-by-step explanation:

Explanation

Parabola:-

The set of points in a plane whose distance from a fixed point and a constant ratio to their corresponding perpendicular distance from a fixed straight line is a conic.

Let S be a fixed point and l be a fixed straight line from any point P,the perpendicular PM is drawn to the line 'l'

The locus of P such that $\frac{SP}{PM} = constant$
The fixed point 'S' is called the Focus.
The fixed line'l 'is called the directrix of the conic
The constant ratio is known as the eccentricity, denoted by 'e'
If e=1 , the conic is called a parabola

1) Step 1 :-

Given the focus S = (2,6) and directrix is x=0

we know that $\frac{SP}{PM}=1$

now cross multiplication , we get

$SP = PM$

squaring on both sides,we get

$SP^{2} = PM^2$

step 2:-

now using distance formula is

$\sqrt(((x_{2}-x_{1})^2+(y_{2} -y_{1} )^2)$

Given S =(2,6) and P(x,y) be any point on parabola

$SP^2 = (x-2)^2+(y-6)^2$ ........(1)

Now using perpendicular distance formula

let P(x , y ) be any point on the parabola

$\frac{ax_{1}+by_{1}+c }{\sqrt{a^2+b^2} }$

Given the directrix is x =0 and P(x,y) be any point on parabola

$PM^2 = \frac{x^2}{\sqrt{1}^2 }$ ......(2)

equating equation(1) and equation (2), on simplification

we get $(x-2)^2+(y-6)^2 = x^2$ .....(3)

apply $(a-b)^2 = a^2+ b^2+2 ab$

now the equation (3) is

$(y-6)^2 = 4 x-4$

now the standard form of parabola is

$(y-k)^2 = 4 a(x-h)$

Final answer:-

$(y-6)^2 = 4 (x-1)$

2) Explanation:-

step 1:

Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)

here the given points of 'x'co- ordinates are equal

Therefore the axis AS is parallel to y- axis

now the standard equation of parabola

$(x-h)^2 = 4 a (y-k)$

now you have to find' a' value

Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)

The distance of $AS = \sqrt{(-5-(-5)^2+(2-6)^2}$

on simplification we get a =4

Final answer :-

the vertex (h,k) = (-5,2) and a=4

$(x-h)^2 = 4 a (y-k)$

The standard parabola is $(x+5)^2 = 16 (y-2)$

5 0

2 years ago

I need help with this

Aleksandr [31]

Answer:

3. m∠1 = 106° ~ this is because ∠1 and ∠2 together make a straight line and are therefore supplementary, meaning added together, they equal 180° (so I did 180° - 74° = 106°)

4. m∠3 = 74° ~ again, it is supplementary to ∠1. It is also equal to ∠2

5. m∠8 = 114° ~ angles opposite of each other (like 1 and 4) are equal (as we know from question 4). From there, we can use the corresponding angle theorem, so we know 4 and 8 are congruent. (also you can just know 1 and 8 are congruent by using the opposite exterior angles theorem)

6. m∠6 = 124° ~ using same-side interior angle theorem, they are supplementary angles (or the corresponding angles theorem mentioned above, make 4 congruent to 8, and 8 is supplementary to 6)

7. m∠7 = 96° ~ using same side exterior angle theorem, these angles are supplementary

8. m∠2 = 64° ~ again, same side exterior angle theorem

4 0

1 year ago