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3 years ago

Find the point on the parabola y^2 = 8x at which the ordinate is double the abscissa.

Mathematics

1 answer:

aliya0001 [1]3 years ago

3 0

Answer:

Step-by-step explanation:

Let's make a complex of functions

y^2 = 8x

y = 2x

Now let's solve it

(2x)^2 = 8x

4x^2 - 8x = 0

4x(x - 2) = 0

x = 0, 2

y = 0, 4

So the answer is (2, 4)

Hope you like it, study well

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Find the value of c, rounded to the nearest tenth A. 11 B. 14.3 C. 37.6 D. 7.4

Helga [31]

Answer:

D. The correct value of c = 7.4

Step-by-step explanation:

According to Tangent-Secant theorem:

"When a tangent and a secant are drawn from one single external point to a circle, square of the length of tangent segment must be equal to the product of lengths of whole secant segment and the exterior portion of secant segment."

Here, the external length of tangent segment = 10

Also, the length of internal segment is 14 and c.

So, by the SECANT THEOREM:

$(10)^2 = 14 \times c\\\implies c = \frac{100}{14} = 7.14$

or, c = 7. 1428

Now, rounding off the value of c = 7. 14 to the nearest tenth, we get

c = 7. 4

Hence the correct value of c = 7.4

8 0

4 years ago

Read 2 more answers

Find the perimeter of J(-4,2), K(1,2), L(2,-2) and M(-3,-2)

Ugo [173]

Answer:

Perimeter = 31.12

Step-by-step explanation:

The perimeter of the shape JKLM will be expressed as;

Perimeter = JK + JL + JM + KL + KM + LM

Using the formula for calculating the distance between two points;

D = √(x2-x1)²+(y2-y1)²

For JK where J(-4,2) and K(1,2)

JK = √(1+4)²+(2-2)²

JK = √5²

JK =√25

JK = 5

For JK where J(-4,2) and L(2,-2)

JL = √(2+4)²+(-2-2)²

JL = √6²+(-4)²

JL =√36+16

JL = √52

For JM where J(-4,2) and M(-3,-2)

JM = √(-3+4)²+(-2-2)²

JM = √1²+(-4)²

JM =√1+16

JM = √17

For KL where K(1,2), L(2,-2)

KL = √(2-1)²+(-2-2)²

KL = √1²+(-4)²

KL =√1+16

KL = √17

For KL where K(1,2), M(-3,-2)

KM = √(-3-1)²+(-2-2)²

KM = √4²+(-4)²

KM = √16+16

KM = √32

KM = 4√2

For KL where L(2,-2) and M(-3,-2)

LM = √(-3-2)²+(-2+2)²

LM = √(-5)²+(0)²

LM = √25

LM = 5

Perimeter = 5 + 4√2 + √17 + √17 + 5 + √52

Perimeter = 10 + 4√2 + 2√17 + √52

Perimeter = 10+5.66+8.25+7.21

Perimeter = 31.12

8 0

3 years ago

Please help me and I’ll give you brainiest!!!!!!!!! Thank you:)!!!!!

sattari [20]

What you have written down is correct, x=-3

4 0

3 years ago

Can 231,000 be written as 0.23

aivan3 [116]

Ur close but it is 0.023 you were off just a bit

3 0

3 years ago

Help with any if you can please thanks.

guajiro [1.7K]

Answer:

1) x= 15

2) x= 34

3) x= 40

Step-by-step explanation:

1) 4x°= 45° +15° (vert. opp angles)

4x°= 60° (simplify)

x°= 60° ÷4 (÷4 on both sides)

x°= 15°

Thus, x= 15.

2) (x+12)° +100° +x°= 180° (adj. angles on a str. line)

2x° +12°= 180° -100°

2x°= 80° -12°

2x°= 68°

x°= 68° ÷2

x°= 34°

x= 34

3) 3x°= 120° (vert. opp. angles)

x°= 120° ÷3

x°= 40°

x= 40

Abbreviations:

1. vert. opp. angles

- vertically opposite angles

- This means that when 2 straight lines intersect, the angles facing opposite each other are equal.

2. adj. angles on a str. line

- adjacent angles on a straight line

- the sum of the angles that lie on a straight line is 180°

4 0

3 years ago

Find the point on the parabola y^2 = 8x at which the ordinate is double the abscissa.​

Find the point on the parabola y^2 = 8x at which the ordinate is double the abscissa.