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Ksju [112]
3 years ago
12

I am sketching parabolas and my book is telling me that I need only two points. But I don't know how I am meant to sketch the pa

rabola accurately with only two points. Are parabolas just supposed to be drawn roughly or do we have to find all the points and then sketch it?
Mathematics
1 answer:
Sever21 [200]3 years ago
7 0

Answer:

In order to sketch a parabola, one will typically use the y-intercept, the x-intercepts, and the vertex.

Step-by-step explanation:

In order to sketch a parabola, one will typically use the given equation in standard form, y = ax^2 + bx + c, and:

  • The y-intercept – the point where the parabola crosses the vertical axis – by substituting 0 for all <em>x</em> values in the given equation and solving for <em>y</em>.
  • The x-intercepts – the points where the parabola crosses the horizontal axis – by setting the given equation equal to 0 (i.e., <em>y</em> = 0) and finding the factors (or roots).
  • The vertex – the point of symmetry where the parabola changes direction and curves up or down – by using x=\frac{-b}{2a} to find the x-coordinate then plugging that value into the given equation to find the y-coordinate.
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Sauron [17]

Answer:

I love algebra anyways

The ans is in the picture with the  steps how i got it

(hope this helps can i plz have brainlist :D hehe)

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
in the equation y equals 8X y is the dependent variable and X is the independent variable what happens to the dependent variable
goblinko [34]
As the independant variable increases, the dependant variable increases at a constant rate creating a linear relationship.
4 0
4 years ago
Given:quadrillateral ABCD inscribed in a circle
son4ous [18]

Answer:

Step-by-step explanation:

Given: quadrilateral ABCD inscribed in a circle

To Prove:

1. ∠A and ∠C are supplementary.

2. ∠B and ∠D are supplementary.

Construction : Join AC and BD.

Proof: As, angle in same segment of circle are equal.Considering AB, BC, CD and DA as Segments, which are inside the circle,

∠1=∠2-----(1)

∠3=∠4-----(2)

∠5=∠6-------(3)

∠7=∠8------(4)

Also, sum of angles of quadrilateral is 360°.

⇒∠A+∠B+∠C+∠D=360°

→→∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360°→→→using 1,2,3,and 4

→→→2∠1+2∠4+2∠6+2∠8=360°

→→→→2( ∠1 +∠6) +2(∠4+∠8)=360°⇒Dividing both sides by 2,

→→→∠B + ∠D=180°as, ∠1 +∠6=∠B , ∠4+∠8=∠B------(A)

As, ∠A+∠B+∠C+∠D=360°

∠A+∠C+180°=360°

∠A+∠C=360°-180°------Using A

∠A+∠C=180°

Hence proved.

credit: someone else

5 0
3 years ago
Line GJ is tangent to point A at point G.<br><br> If AG = 9 and GJ = 12, find AJ.
kompoz [17]
GJ is a tangent
АG is a radius
A tangent line to a circle is perpendicular to the radius drawn to the tangent point   ⇒ m∠АGJ = 90°  ⇒ ΔАGJ is the right triangle.

<span>By the Pythagorean theorem:
</span>AJ² = АG² + GJ²
AJ² = 9² + 12² = 81 + 144 = 225
AJ = √225 = 15 units.
4 0
3 years ago
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Please help!! <br>circles combos are too hard​
malfutka [58]

Answer:

56 + 53pi

Step-by-step explanation:

<u><em>Area of small circles:</em></u>

diameter of small circle: 4cm

forumla to find area of circle: A = pir^2

r is radius = half of diameter -> d/2 = 4 / 2 = 2cm

A = pi (2cm)^2

A = pi (4cm)

A = 4pi

<u><em>Area of large circle:</em></u>

diameter of small circle: 4cm

forumla to find area of circle: A = pir^2

r is radius = half of diameter -> d/2 = 14 / 2 = 7cm

A = pi (7cm)^2

A = pi (49cm)

A = 49pi

<u><em>Area of rectangle:</em></u>

Area = width x length

Area = 14cm x 4cm

Area = 56cm

<u><em>Add all three areas:</em></u>

Area of rectangle + large circle + small circle

56cm + 49pi + 4pi = 56cm + 53pi

7 0
2 years ago
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