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LekaFEV [45]
3 years ago
12

20 points. MAY YOU HELP FAST PLEASE? The thing ends tonight.

Mathematics
1 answer:
deff fn [24]3 years ago
7 0

2) cant help here

3) b

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PLZ HELP AND HURRY!!!!
yuradex [85]

Answer: x=30, Good Luck!

Step-by-step explanation:

45/27 = 1.6666666...

1.66666...(18) = 30

8 0
3 years ago
In x² , what is the 2 called?
alexandr402 [8]

Answer:

<h2>an exponent</h2>

Step-by-step explanation:

In x², here the 2 is exponent while x is base.

7 0
3 years ago
Read 2 more answers
Use the parabola tool to graph the quadratic function f(x)=x2−12x+27. Graph the parabola by first plotting its vertex and then p
alina1380 [7]

Answer:

vertex is (6 , -9)

points are (9,0) and (3,0)

Step-by-step explanation:

Given quadratic function f(x)=x^2-12x+27

We have to plot the given quadratic function.

Consider the Given quadratic function f(x)=x^2-12x+27

The general form of quadratic function  is given f(x)=a(x-h)^2+k

Where, (h, k)  is vertex , given  h =\frac{-b}{2a} and k = f(h)  

Thus, for given quadratic function f(x)=x^2-12x+27

a = 1 , b= -12 , c = 27

Thus,

h =\frac{-b}{2a}=\frac{12}{2}=6

k = f(h) that is f(12) = (6)^2 - 12× 6 +27 = 36 - 72 + 27 = - 9

Thus, given  quadratic function f(x)=x^2-12x+27 in standard form is f(x)=(x-6)^2-9  

Thus, vertex is (6 , -9)

For second point put f(x)=0  , we get,

f(x)=(x-6)^2-9=0  

\Rightarrow (x-6)^2-9=0  

\Rightarrow (x-6)^2=9  

\Rightarrow (x-6)=\pm 3  

\Rightarrow x= 6\pm 3  

Thus, \Rightarrow x=6+3=9 and \Rightarrow x=6-3=3  

thus, points are (9,0) and (3,0)

Graph is attached below.

5 0
4 years ago
Jacob has some cookies. 3/4 of his cookies are chocolate chip. Out of the chocolate chip cookies 1/8 of them have nuts as well.
Tatiana [17]
Well, we can find the answer by multiplying the two fractions. 3/4*1/8 equals 3/32.

3 0
4 years ago
Read 2 more answers
Write the equation of the circle in general form. Show your work.
vlabodo [156]
(x+1)²+(y+1)²=9

the answer can be found with 
(x-h)²+(y+k)²=r²
where the point (h,k) is the center.
7 0
4 years ago
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