Subtract: (3x2 − 8x + 3) − (−5x2 + 4x − 5)

valkas [14]

Answer:

Ray. Definition: A portion of a line which starts at a point and goes off in a particular direction to infinity. Try this Adjust the ray below by dragging an orange dot and see how the ray AB behaves. Point A is the ray's endpoint. -Google

Step-by-step explanation:

7 0

3 years ago

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A survey handed out to individuals at a major shopping mall in a small Florida city in July asked the following: Gender Age Ethn

AysviL [449]

Answer and Explanation:

Gender data is categorical as it can be sorted into two categories - males and females - and does not involve any numerical values.

Age data is ratio as each person's age is a numerical value and it has a natural zero as point of origin. Moreover, the numerical values cannot be negative.

Ethnicity data is categorical as the various ethnicities to which those surveyed can belong are separate categories. There are no numerical values involved.

Length of residency data is ratio as the length duration is measured in number of years, thus having a numerical value. There is also a natural zero as point of origin and no possible negative values.

Overall satisfaction with city services data is ordinal as it is based on a ranking from poor to excellent.

Quality of schools data is also ordinal as it involves a ranking or definite ordering of data - from poor to excellent.

4 0

2 years ago

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

Help - Math Test-need answer immediately I’ll give brainliest.

ludmilkaskok [199]

Answer:

The answer is C the 2 rectangles

5 0

2 years ago

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Suppose the roots of the polynomial $x^2 - mx + n$ are positive prime integers (not necessarily distinct). Given that $m < 20

Vsevolod [243]

Answer:

18 values for n are possible.

Step-by-step explanation:

Given the quadratic polynomial:

$x^2 - mx + n$

such that:

Roots are positive prime integers and

$m < 20$

To find:

How many possible values of $n$ are there ?

Solution:

First of all, let us have a look at the sum and product of a quadratic equation.

If the quadratic equation is:

$Ax^{2} +Bx+C$

and the roots are: $\alpha$ and $\beta$

Then sum of roots, $\alpha+\beta = -\frac{B}{A}$

Product of roots, $\alpha \beta = \frac{C}{A}$

Comparing the given equation with standard equation, we get:

A = 1, B = -m and C = n

Sum of roots, $\alpha+\beta = -\frac{-m}{1} = m$

Product of roots, $\alpha \beta = \frac{n}{1} = n$

We are given that $m$

$\alpha$ and $\beta$ are positive prime integers such that their sum is less than 20.

Let us have a look at some of the positive prime integers:

2, 3, 5, 7, 11, 13, 17, 23, 29, .....

Now, we have to choose two such prime integers from above list such that their sum is less than 20 and the roots can be repetitive as well.

So, possible combinations and possible value of $n (= \alpha \times \beta)$ are:

$1.\ 2, 2\Rightarrow n = 2\times 2 = 4\\2.\ 2, 3 \Rightarrow n = 6\\3.\ 2, 5 \Rightarrow n = 10\\4.\ 2, 7\Rightarrow n = 14\\5.\ 2, 11 \Rightarrow n = 22\\6.\ 2, 13 \Rightarrow n = 26\\7.\ 2, 17 \Rightarrow n = 34\\8.\ 3, 3\Rightarrow n = 3\times 3 = 9\\9.\ 3, 5 \Rightarrow n = 15\\10.\ 3, 7 \Rightarrow n = 21\\$

$11.\ 3, 11\Rightarrow n = 33\\12.\ 3, 13 \Rightarrow n = 39\\13.\ 5, 5 \Rightarrow n = 25\\14.\ 5, 7 \Rightarrow n = 35\\15.\ 5, 11 \Rightarrow n = 55\\16.\ 5, 13 \Rightarrow n = 65\\17.\ 7, 7 \Rightarrow n = 49\\18.\ 7, 11 \Rightarrow n = 77$

So,as shown above 18 values for n are possible.

3 0

3 years ago