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Zepler [3.9K]
4 years ago
9

My dividend is 24. I am 2 more than my quotient. I have no remainder. what divisor am i?

Mathematics
2 answers:
Virty [35]4 years ago
6 0

Answer:

The divisor can be 6 or -4.

Step-by-step explanation:

Dividend = Divisor × Quotient + Remainder

Let the quotient be x.

The number or divisor = x +2

Remainder = 0

Dividend = 24

24=(x+2)\times x+0

x^2+2x-24=0

Using middle term splitting :

x^2+6x-4x-24=0

x(x+6)-4(x+6)=0

(x-4)(x+6)= 0

if x = 4

Then ,Divisor = 4 + 2 = 6

if x = -6

Then ,Divisor = -6 + 2 = -4

The divisor can be 6 or -4.

andreev551 [17]4 years ago
3 0
Divident =divisor × quotient + remainder
24= divisor ×(divisor-2)+0
d^2-2d-24=0
d=6 or -4
quotient is 4 of -6
so there are two possibilities, 6 or -4
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A parabola can be drawn given a focus of (4, -3) and a directrix of y=5 Write the equation of the parabola in any form.
Serggg [28]

Answer:

The equation of the parabola is (x - 4)² = -16(y - 1)

Step-by-step explanation:

The standard form of the equation of the parabola is (x - h)² = 4p(y - k), where

  • The vertex of the parabola is (h, k)
  • The focus is (h, k + p)
  • The directrix is at y = k - p  

∵ The focus of the parabola is (4, -3)

∵ The focus is (h, k + p)

∴ h = 4

∴ k + p = -3 ⇒ (1)

∵ It has a directrix of y = 5

∵ The directrix of the parabola is y = k - p

∴ k - p = 5 ⇒ (2)

→ Add equations (1) and (2) to find k and p

∵ (k + k) + (p - p) = (-3 + 5)

∴ 2k + 0 = 2

∴ 2k = 2

→ Divide both sides by 2

∴ k = 1

→ Substitute the value of k in equation (1)

∵ 1 + p = -3

→ Subtract 1 from both sides

∴ 1 - 1 + p = -3 - 1

∴ p = -4

∵ The form of the equation of the parabola is (x - h)² = 4p(y - k)

→ Substitute the values of h, k, p in it

∴ (x - 4)² = 4(-4)(y - 1)

∴ (x - 4)² = -16(y - 1)

∴ The equation of the parabola is (x - 4)² = -16(y - 1)

3 0
3 years ago
Fit a quadratic function to these three points: (-1,-11) , (0,-3) and (3,-27)
notka56 [123]
It is -10 then -3 then -24

6 0
3 years ago
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harkovskaia [24]
1. The midpoint of the segment joining points (a, b) and ( j, k) is ((j+a)/2,(k+b)/2)

2. Let the coordinate of H be (a, b)
T(0, 4) = ((a + 0)/2, (b + 2)/2)
(a + 0)/2 = 0 => a + 0 = 0 => a = 0
(b + 2)/2 = 4 => b + 2 = (2 x 4) = 8 => b = 8 - 2 = 6
Therefore, the cordinate of H is (0, 6)

3. Point (-4, 3) lies in Quadrant II

4. Point (6, 0) lies on the x-axis

5. Any line with no slope is parallel to the y-axis

7. a is the value of the x-coordinate.
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5a = 8 - 3 = 5
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8. Equation of a circle is given by (x - a)^2 + (y - b)^2 = r^2; where (a, b) is the center and r is the radius.
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For the given circle (x + 5)^2 + (y - 7)^2 = 36 => (x - (-5))^2 + (y - 7)^2 = 6^2 => r = 6.
Therefore, its radius is 6

10. If the equation of a circle is (x - 2)^2 + (y - 6)^2 = 4, the center is point (2, 6).
True

7 0
4 years ago
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Suppose that a sphere with radius 5a has the same volume as a cone of radius 3a. What is the height of the cone? Give your answe
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(4/3) pi 125a^3 = (1/3) pi 9a^2 h

(4/3) * 125a^3 = (1/3) 9a^2 h

h  =  (4/3) * 125 a^3
        --------------------  = 4 * 13.889 a
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h =  55.56 cu units  to nearest  1/100
4 0
3 years ago
.......................................
klio [65]

Answer:

Step-by-step explanation:

d

7 0
3 years ago
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