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wolverine [178]
3 years ago
9

Which one is larger and why 12 x 3/7 or 6 x 3/7

Mathematics
1 answer:
Serggg [28]3 years ago
6 0

Answer:12 x 3/7 is larger than 6 x 3/7.

12x3/7=36/7 while 6x3/7=18/7 since both fractions 36/7 and 18/7 have the same denominators, to know which is larger we just check the numerator,the fraction with the largest numerator is largest which is 36/7

Step-by-step explanation:

12x3/7=(12x3)/7=36/7

6x3/7=(6x3)/7=18/7

36/7 is larger than 18/7

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What types of erosion creates caves?
pishuonlain [190]
Caves are formed by the dissolution of limestone. Rainwater picks up carbon dioxide from the air and as it percolates through the soil, which turns into a weak acid. This slowly dissolves out the limestone along the joints, bedding planes and fractures, some of which become enlarged enough to form caves.
4 0
3 years ago
What is the center and radius of -10x = -121 - y2 - x2 -22y
OlgaM077 [116]

We need to find the center and the radius of

-10x\: =\: -121\: -\: y^2\: -\: x^2\: -22y

The general circle equation is the following

(x-h)^2+(y-k)^2=r^2

where

(h,k) is the center and

r is the radius

1. rearrange the equation

(x^2-10x)+(y^2+22y+121)=0

2. Add 25 on both sides

(x^2-10x+25)+(y^2+22y+121)=25

3. Factor

(x-5)^2+(y+11)^2=5^2

Now we have an equation that is very similar to the circle equation, so let's compare them

Center -> (h,k) = (5,-11)

radius -> r = 5

8 0
1 year ago
Find the nature of the root
GREYUIT [131]

Answer:

1) Real and same.

2) Real and distinct

3) Real and distinct

4) Real and distinct

5) Real and distinct

6) Real and distinct

7) Real and distinct

8) Real and distinct

Step-by-step explanation:

If a quadratic equation is in the form of ax² + bx + c = 0, then Discriminant of the equation D = b² - 4ac, which governs the nature of roots the equation has.

If D > 0, then there will be two different real roots.

If D = 0, then two same and real roots.

If D < 0, then two distinct but imaginary roots.

Now, 1) x² + 6x + 9 = 0 has D = 6² - 4 × 9 × 1 = 0. So, there will be two same and real roots.

2) 5x² - x = 4x² + 2

⇒ x² - x - 2 = 0

It has D = (-1)² - 4 × 1 × (-2) = 9. Therefore, the roots will be two distinct and real.

3) \frac{2}{x} + \frac{3}{x} = x - 4

⇒ 5 = x² - 4x

⇒ x² - 4x - 5 = 0.

So, D = (-4)² - 4 × (1) × (- 5) = 36. So, the equation has two real and distinct rools.

4) x(x - 5) = 4(5 - x)

⇒ x² - x - 20 = 0

Hence, D = (-1)² - 4 × 1 × (-20) = 81

So, the roots will be real and distinct.

5) x² + 7x + 1 = 0 has D = 7² - 4 × 1 × 1 = 45

So, the roots will be real and distinct.

6) 2x² + 9x + 3 = 0, has D = 9² - 4 × 2 × 3 = 57.

Hence, the roots will be real and distinct.

7) 5x² - 6 = 13x

⇒ 5x² - 13x - 6 = 0

So, D = (-13)² - 4 × 5 × (-6) = 289

So, the roots will be real and distinct.

8) x² - x = 3(x + 7)

⇒ x² - 4x - 21 = 0

It has D = (-4)² - 4 × 1 × (-21) = 100

So, the roots will be real and distinct. (Answer)

6 0
3 years ago
Help me on these questions
mojhsa [17]

Answer:

a) The equation is (y - 1)² = -8 (x - 4)

b) The equation is (x - 1)²/25 + (y - 4)²/16 = 1

c) The equation of the ellipse is (x - 3)²/16 + y²/4 = 1

Step-by-step explanation:

a) Lets revise the standard form of the equation of the parabola with a

   horizontal axis

# (y - k)² = 4p (x - h), (h , k) are the coordinates of its vertex and p ≠ 0

- The focus of it is (h + p , k)

* Lets solve the problem

∵ The focus is (2 , 1)

∵ focus is (h + p , k)

∴ h + p = 2 ⇒ subtract p from both sides

∴ h = 2 - p ⇒ (1)

∴ k = 1

∵ It opens left, then the axis is horizontal and p is negative

∴ Its equation is (y - k)² = 4p (x - h)

∵ k = 1

∴ Its equation is (y - 1)² = 4p (x - h)

- The parabola contains point (2 , 5), substitute the coordinates of the

 point in the equation of the parabola

∴ (5 - 1)² = 4p (2 - h)

∴ (4)² = 4p (2 - h)

∴ 16 = 4p (2 - h) ⇒ divide both sides by 4

∴ 4 = p (2 - h) ⇒ (2)

- Use equation (1) to substitute h in equation (2)

∴ 4 = p (2 - [2 - p]) ⇒ open the inside bracket

∴ 4 = p (2 - 2 + p) ⇒ simplify

∴ 4 = p (p)

∴ 4 = p² ⇒ take √ for both sides

∴ p = ± 2, we will chose p = -2 because the parabola opens left

- Substitute the value of p in (1) to find h

∵ h = 2 - p

∵ p = -2

∴ h = 2 - (-2) = 2 + 2 = 4

∴ The equation of the parabola in standard form is

  (y - 1)² = 4(-2) (x - 4)

∴ The equation is (y - 1)² = -8 (x - 4)

b) Lets revise the equation of the ellipse

- The standard form of the equation of an ellipse with  center (h , k)

 and major axis parallel to x-axis is (x - h)²/a² + (y - k)²/b² = 1  

- The coordinates of the vertices are (h ± a , k )  

- The coordinates of the foci are (h ± c , k), where c² = a² - b²  

* Now lets solve the problem

∵ Its vertices are (-4 , 4) and (6 , 4)

∵ The coordinates of the vertices are (h + a , k ) and (h - a , k)  

∴ k = 4

∴ h + a = 6 ⇒ (1)

∴ h - a = -4 ⇒ (2)

- Add (1) and (2) to find h

∴ 2h = 2 ⇒ divide both sides by 2

∴ h = 1

- Substitute the value of h in (1) or (2) to find a

∴ 1 + a = 6 ⇒subtract 1 from both sides

∴ a = 5

∵ The foci at (-2 , 4) and (4 , 4)

∵ The coordinates of the foci are (h + c , k) , (h - c , k)

∴ h + c = 4

∵ h = 1

∴ 1 + c = 4 ⇒ subtract 1 from both sides

∴ c = 3

∵ c² = a² - b²

∴ 3² = 5² - b²

∴ 9 = 25 - b² ⇒ subtract 25 from both sides

∴ -16 = -b² ⇒ multiply both sides by -1

∴ 16 = b²

∵ a² = 25

∵ The equation of the ellipse is (x - h)²/a² + (y - k)²/b² = 1

∴ The equation is (x - 1)²/25 + (y - 4)²/16 = 1

c) How to identify the type of the conic  

- Rewrite the equation in the general form,  

 Ax² + Bxy + Cy² + Dx + Ey + F = 0  

- Identify the values of A and C from the general form.  

- If A and C are nonzero, have the same sign, and are not equal  

 to each other, then the graph is an ellipse.  

- If A and C are equal and nonzero and have the same sign, then

 the graph is a circle  

- If A and C are nonzero and have opposite signs, and are not equal  

 then the graph is a hyperbola.  

- If either A or C is zero, then the graph is a parabola  

* Now lets solve the problem

∵ x² + 4y² - 6x - 7 = 0

∵ The general form of the conic equation is

   Ax² + Bxy + Cy² + Dx + Ey + F = 0  

∴ A = 1 and C = 4

∵ If A and C are nonzero, have the same sign, and are not equal  to

  each other, then the graph is an ellipse.

∵ x² + 4y² - 6x - 7 = 0 ⇒ re-arrange the terms

∴ (x² - 6x ) + 4y² - 7 = 0

- Lets make x² - 6x completing square

∵ 6x ÷ 2 = 3x

∵ 3x = x × 3

- Lets add and subtract 9 to x² - 6x to make the completing square

 x² - 6x + 9 = (x - 3)²

∴ (x² - 6x + 9) - 9 + 4y² - 7 = 0 ⇒ simplify

∴ (x - 3)² + 4y² - 16 = 0 ⇒ add 16 to both sides

∴ (x - 3)² + 4y² = 16 ⇒ divide all terms by 16

∴ (x - 3)²/16 + 4y²/16 = 1 ⇒ simplify

∴ (x - 3)²/16 + y²/4 = 1

∴ The equation of the ellipse is (x - 3)²/16 + y²/4 = 1

5 0
3 years ago
Help please Which number sentence below matches this word problem?
Ostrovityanka [42]

Answer:

I think it's d

Step-by-step explanation:

Nit 100% sure but if nit its b

4 0
2 years ago
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