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vladimir2022 [97]
3 years ago
8

PLEASE HELP I'll give extra POINTS.

Mathematics
1 answer:
lys-0071 [83]3 years ago
5 0
Is there a picture for this?
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What is the area of the model in the problem?
nata0808 [166]

Answer:

A=x^2+12x+27\ units^2

Step-by-step explanation:

we know that

The area of the model is equal to the area of a rectangle

The area of a rectangle is equal to

A=LW

we have

L=x+9\ units

W=x+3\ units

substitute

A=(x+9)(x+3)

Apply distributive property

A=x^2+3x+9x+27

Combine like terms

A=x^2+12x+27\ units^2

4 0
4 years ago
Cuáles son todos los factores de 30​
Novay_Z [31]

Answer:

Los factores de 30 son: 1 y 30; 2 y 15; 3 y 10; 5 y 6.

Step-by-step explanation:

6 0
3 years ago
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11. A set of weights includes a 5 lb bar and 8 pairs of weight plates. Each pair of plates weighs 10 lbs. If x
Marat540 [252]
X is amount of plates 10lbs and f(x) is the function
4 0
3 years ago
SEE PICTURE FOR QUESTION​
castortr0y [4]
I think the answer is B
7 0
3 years ago
(03.09 MC) Derive the equation of the parabola with a focus at (−2, 4) and a directrix of y = 6. Put the equation in standard fo
Nookie1986 [14]
Check the picture below.

so the focus point is at -2, 4 and the directrix is at y = 6, now, keeping in mind that the vertex is half-way between those two fellows, from 4 to 6, it'd be the y-coordinate of 5, and therefore, the vertex is at -2,5, as you see there in the picture, and the parabola looks like so.  Since the parabola is a vertical one, the squared variable is the "x".

notice the distance "p", is just 1 unit, however, since the parabola is opening downwards, "p" is negative, and thus -1.

\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
4p(x- h)=(y- k)^2
\\\\
\boxed{4p(y- k)=(x- h)^2}
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ( h, k)\\\\
 p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
h=-2\\
k=5\\
p=-1
\end{cases}\implies 4(-1)(y-5)=[x-(-2)]^2
\\\\\\
-4(y-5)=(x+2)^2\implies y-5=-\cfrac{1}{4}(x+2)^2
\\\\\\
y=-\cfrac{1}{4}(x+2)^2+5

5 0
3 years ago
Read 2 more answers
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