How would you write 834,209 in word form

Use numerals instead of words. If necessary, use / for the fraction bar(s). A parabola is given by the equation y = x2 + 4x + 4.

Anarel [89]

Graph x2 = 4y and state the vertex, focus, axis of symmetry, and directrix.This is the same graphing that I've done in the past: y = (1/4)x2. So I'll do the graph as usual: The vertex is obviously at the origin, but I need to "show" this "algebraically" by rearranging the given equation into the conics form:x2 = 4y Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved
(x – 0)2 = 4(y – 0)This rearrangement "shows" that the vertex is at (h, k) = (0, 0). The axis of symmetry is the vertical line right through the vertex: x = 0. (I can always check my graph, if I'm not sure about this.) The focus is "p" units from the vertex. Since the focus is "inside" the parabola and since this is a "right side up" graph, the focus has to be above the vertex.From the conics form of the equation, shown above, I look at what's multiplied on the unsquaredpart and see that 4p = 4, so p = 1. Then the focus is one unit above the vertex, at (0, 1), and the directrix is the horizontal line y = –1, one unit below the vertex.vertex: (0, 0); focus: (0, 1); axis of symmetry: x = 0; directrix: y = –1Graph y2 + 10y + x + 25 = 0, and state the vertex, focus, axis of symmetry, and directrix.Since the y is squared in this equation, rather than the x, then this is a "sideways" parabola. To graph, I'll do my T-chart backwards, picking y-values first and then finding the corresponding x-values for x = –y2 – 10y – 25:To convert the equation into conics form and find the exact vertex, etc, I'll need to convert the equation to perfect-square form. In this case, the squared side is already a perfect square, so:y2 + 10y + 25 = –x 
(y + 5)2 = –1(x – 0)This tells me that 4p = –1, so p = –1/4. Since the parabola opens to the left, then the focus is 1/4 units to the left of the vertex. I can see from the equation above that the vertex is at (h, k) = (0, –5), so then the focus must be at (–1/4, –5). The parabola is sideways, so the axis of symmetry is, too. The directrix, being perpendicular to the axis of symmetry, is then vertical, and is 1/4 units to the right of the vertex. Putting this all together, I get:vertex: (0, –5); focus: (–1/4, –5); axis of symmetry: y = –5; directrix: x = 1/4Find the vertex and focus of y2 + 6y + 12x – 15 = 0The y part is squared, so this is a sideways parabola. I'll get the y stuff by itself on one side of the equation, and then complete the square to convert this to conics form.y2 + 6y – 15 = –12x 
y2 + 6y + 9 – 15 = –12x + 9 
(y + 3)2 – 15 = –12x + 9 
(y + 3)2 = –12x + 9 + 15 = –12x + 24 
(y + 3)2 = –12(x – 2) 
(y – (–3))2 = 4(–3)(x – 2)Then the vertex is at (h, k) = (2, –3) and the value of p is –3. Since y is squared and p is negative, then this is a sideways parabola that opens to the left. This puts the focus 3 units to the left of the vertex.vertex: (2, –3); focus: (–1, –3)

6 0

3 years ago

Can anybody help me please hurry

zhuklara [117]

Yoo kid for these type of problems your better o go with an app called Photomath

8 0

3 years ago

8-10y+12x-15xy factorise

Alik [6]

Answer:

(4 - 5y)(2 + 3x)

Step-by-step explanation:

Given

8 - 10y + 12x - 15xy ( factor the first/second and third/fourth terms )

= 2(4 - 5y) + 3x(4 - 5y) ← factor out (4 - 5y) from each term

= (4 - 5y)(2 + 3x) ← in factored form

3 0

3 years ago

Please help me understand this problem! It would be greatly appreciated <3 much love

cricket20 [7]

Answer:

all work is shown and pictured

3 0

4 years ago

Give the expression 4n^2t/2n^t-1, write the expression as a binomial.

lubasha [3.4K]

binomial is an algebraic expression containing 2 terms. For example, (x + y) is a binomial.

We sometimes need to expand binomials as follows:

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

Clearly, doing this by direct multiplication gets quite tedious and can be rather difficult for larger powers or more complicated expressions.

Pascal's Triangle

We note that the coefficients (the numbers in front of each term) follow a pattern. [This was noticed long before Pascal, by the Chinese.]

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

You can use this pattern to form the coefficients, rather than multiply everything out as we did above.

The Binomial Theorem

We use the binomial theorem to help us expand binomials to any given power without direct multiplication. As we have seen, multiplication can be time-consuming or even not possible in some cases.

Properties of the Binomial Expansion (a + b)nThere are \displaystyle{n}+{1}n+1 terms.The first term is an and the final term is bn.Progressing from the first term to the last, the exponent of a decreases by \displaystyle{1}1 from term to term while the exponent of b increases by \displaystyle{1}1. In addition, the sum of the exponents of a and b in each term is n.If the coefficient of each term is multiplied by the exponent of a in that term, and the product is divided by the number of that term, we obtain the coefficient of the next term.

7 0

3 years ago