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ahrayia [7]
3 years ago
7

create a perfect square trinomial that can be factored using special patterns. Factor the trinomial and show all work for full c

redit.
Mathematics
2 answers:
Ulleksa [173]3 years ago
8 0
Trinomial is 3 terms
ax^2+bx+c
4 is a perfect square since 2*2
essentially, you need (x+a)^2
(x+1)^2=(x+1)(x+1)=x^2+2x+1
as far as "special patterns" this may imply that you need to find a perfect square trinomial that requires a different factoring process, I'm not sure what that means
Lyrx [107]3 years ago
3 0

Answer: 4x^2+12x+9


Step-by-step explanation:

A perfect square trinomial is written as ax^2+bx+c, where

first term ax^2 = square of first term of binomial

second term=bx=twice the product of both terms of binomial.

and third term 'c'=square of last term of binomial

Thus to create a perfect  square trinomial put 'a' and 'c' a square number

Let a=4 and c=9

The required trinomial will be

4x^2+12x+9

=(2x)^2+2(2x)(3)+3^2\\=(2x+3)^2.......\text{[using pattern}(a+b)^2=a^2+2ab+b^2]\\=(2x+3)(2x+3)

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Is 7900 100 times as much as 79
Cloud [144]

Answer:

Yes

Step-by-step explanation:

4 0
3 years ago
Find two square numbers that have a sum of 130
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7 squared = 49 and 9 squared= 81 add up 49 and 81 and you get 130
8 0
4 years ago
Solve this for me please ? I do not know the steps or answer
yan [13]
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3 0
4 years ago
How do I use intercepts to graph 3y= - 5x - 30
expeople1 [14]

Answer:

y-intercept is (0,-10) and x-intercept is (-6,0).  Connect them by a straight line to graph the given equation.

Step-by-step explanation:

The given equation of line is

3y=-5x-30

For x=0,

3y=-5(0)-30

3y=-30

y=-10

So, y-intercept is at point (0,-10).

For y=0,

3(0)=-5x-30

0=-5x-30

5x=-30

x=-6

So, x-intercept is at point (-6,0).

Now, plot the point (0,-10) and (-6,0) on a coordinate plane and connect them by a straight line to graph the given line as shown below.

7 0
3 years ago
HELP<br> Use the definition of continuity and the properties of limits to show that’s the function…
Solnce55 [7]

Answer with Step-by-step explanation:

We are given that

g(x)=(x^2-3x-10)(x^2+2x^2-5x+4)

x=-1

Using property of limit

\lim_{x\rightarrow -1}g(x)=\lim_{x\rightarrow -1}(x^2-3x-10)(3x^2-5x+4)

=(1+3-10)(3+5+4)

=(-6)(12)

=-72

g(-1)=(1+3-10)(1+2+5+4)

g(-1)=(-6)(12)=-72

\lim_{x\rightarrow -1}g(x)=g(-1)

Hence, the function is continuous at x=-1

4 0
3 years ago
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