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3 years ago

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Which of the following are characteristics of the graph of the quadratic parent function? Check all that apply. A. It is a parab

ola. B. It is in quadrants I and II. C. It is a straight line. D. It is in quadrants I and III.

2 answers:

jolli1 [7]3 years ago

7 0

Answer – A and B
A. It is a parabola
B. It is in quadrants I and II
The most simple quadratic function is y = x^2. The graph drawn for this function, y = x^2) is known as the graph of the quadratic parent function OR the parent function for parabolas. This graph has some few characteristics:
- It is the simplest parabola (Generally, the graph of any quadratic function is a parabola).
- It passes through the origin (0,0).
- It is contained in Quadrants I and II. <span>
</span>

MaRussiya [10]3 years ago

3 0

Answer:

The correct options are :

A. It is a parabola.

B. It is in quadrants I and II.

Step-by-step explanation:

To check the characteristics of the quadratic parent function, let us take the basic quadratic function : y = x²

The graph for this parent quadratic f unction is attached below :

Now, as seen from the graph below :

It is basic parabola. And generally the graphs of the quadratic function are in the form of parabola.

The graph lies in the I and II quadrants

So. these can only be the characteristics of the graph of the quadratic parent function.

Hence, The correct options are :

A. It is a parabola.

B. It is in quadrants I and II.

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The area of polygon MNOPQR = Area of a rectangle that is 9 square units + Area of a rectangle that is ___ square units. (Input w

vekshin1

You have the polygon MNOPQR which can be expressed as two rectangles pasted one next to each other.

To see the two rectangles in the picture, you can draw a line parallel to segment MR througn point N.

From the original picture you can state the dimensions of both rectangles.

Call S, the point where the line that you drew intercepts the segment RQ.

Then one rectangle is MNSR and the other rectangle is OPQS.

The measures of the sides of the rectangle MNSR are:

- the length of MN = length of SR = base

- the length of MR = the length of NS.= height

So its area is base * height, which you can all A1.

The measured of the rectangle OPQS are:

- segment OP = segment SQ = segment QR - segment SR = base

- segment PQ = segment OS = height

So its area is base * height, which you can call A2.

Then the area of the polygon MNOPQRS is A1 + A2. One of them is 9 u^2 and the other is what the answer is asking for, and that you have calculated above.

With this procedure you can tell the value needed.

8 0

3 years ago

Find the distance between (-5,6)and (3,2).

skad [1K]

Answer:

$\displaystyle d = 4\sqrt{5}$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Algebra I</u>

Coordinates (x, y)

<u>Algebra II</u>

Distance Formula: $\displaystyle d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Step-by-step explanation:

<u>Step 1: Define</u>

Point (-5, 6) → x₁ = -5, y₁ = 6

Point (3, 2) → x₂ = 3, y₂ = 2

<u>Step 2: Find distance </u><em><u>d</u></em>

Simply plug in the 2 coordinates into the distance formula to find distance <em>d</em>

Substitute in points [Distance Formulas]: $\displaystyle d = \sqrt{(3+5)^2+(2-6)^2}$
[Distance] [√Radical] (Parenthesis) Add/Subtract: $\displaystyle d = \sqrt{(8)^2+(-4)^2}$
[Distance] [√Radical] Evaluate exponents: $\displaystyle d = \sqrt{64+16}$
[Distance] [√Radical] Add: $\displaystyle d = \sqrt{80}$
[Distance] [√Radical] Simplify: $\displaystyle d = 4\sqrt{5}$

6 0

3 years ago

Find the dicontinuities of the function.

elena-s [515]

Answer:

Step-by-step explanation:

I'm assuming you meant to type in

$f(x)=\frac{x^2+12x+27}{x^2+4x+3}$ because you can only have removable discontinuities where there is a rational (fraction) function. Begin by factoring both the numerator and denominator to

$f(x)=\frac{(x+3)(x+9)}{(x+1)(x+3)}$ and cancelling out like terms would have us eliminating the (x + 3). That is where there is a removable discontinuity. It leaves a hole. The other discontinuity, (x + 1) doesn't cancel out so it is a non-removable discontuinity, which is a vertical asymptote.

The removable discontinuity is at -3. There is no y value at x = -3 (remember there's only a hole here), because -3 causes the denominator to go to 0 and we all know that having a 0 in the denominator of a fraction is a big no-no!!!

5 0

3 years ago

A gas container in the shape of a cone has a height of 12 inches and a maximum diameter of 8 inches. When the container is fille

Nady [450]

Answer:

150.9 cubic inches

Step-by-step explanation:

volume of he gas = volume of a cone x 3/4

Volume of a cone 1/3(nr^2h)

n = 22/7

r = radius

h = height

the diameter is the straight line that passes through the centre of a circle and touches the two edges of the circle.

A radius is half of the diameter

8/2 = 4 inches

1/3 (22/7 x 4^2 x 12) = 201.142857 in^3

volume of the gas = 3/4 x 201.142857 in^3 = 150.9 in^3

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3 0

3 years ago

Given right triangle ABCABC with altitude \overline{BD} BD drawn to hypotenuse ACAC. If BC=10BC=10 and DC=5,DC=5, what is the le

Helen [10]

Answer: $\overline{AC}=$ 12.5

Step-by-step explanation: The triangle ABC and its altitude $\overline{BD}$ is represented in the figure below.

<u>Altitude</u> is a segment of line that link a vertex and the opposite side, forming a right angle.

So, because of $\overline{BD}$ , now we have two similar triangles, which means that ratios of corresponding sides are equal:

$\frac{\overline{BD}}{\overline{BC}} =\frac{\overline{AD}}{\overline{BD}}$

$\overline{BD}^{2}=\overline{BC}.\overline{AD}$ (1)

This is always true for a right triangle and a altitude drawn to the hypotenuse.

Triangle BDC is also right triangle. So, we can use Pythagorean theorem to determine the missing side.

$\overline{BC}^{2}=\overline{CD}^{2}+\overline{BD}^{2}$

$\overline{BD}^{2}=\overline{BC}^{2}-\overline{CD}^{2}$ (2)

Substituting (2) into (1):

$\overline{BC}^{2}-\overline{CD}^{2}=\overline{BC}.\overline{AD}$

We want to find f, so:

$\overline{AD}=\frac{\overline{BC}^{2}-\overline{CD}^{2}}{\overline{BC}}$

$f=\frac{10^{2}-5^{2}}{10}$

f = 7.5

The length of $\overline{AC}$ is

$\overline{AC}=f+5$

$\overline{AC}=7.5+5$

$\overline{AC}=12.5$

The length of the hypotenuse of triangle ABC is 12.5 units.

5 0

3 years ago

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