In standard form: (x+5)(3x-1)-(x-4)^2

Mathematics

2 answers:

denpristay [2]3 years ago

6 0

I hope this helps you

x.3x+x.(-1)+5.3x+5.(-1)-(x^2-2.4.x+16)

3x^2-x+15x-5-x^2+8x-16

2x^2+22x-21

Minchanka [31]3 years ago

5 0

Basically, expand
expand them seperately to reduce confusion
exponent first
(x-4)^2=(x-4)(x-4)=x²-4x-4x+16=x²-8x+16

(x+5)(3x-1)=3x²+15x-x-5=3x²+14x-5

3x²+14x-5-(x²-8x+16)
3x²+14x-5-x²+8x-16
3x²-x²+14x+8x-5-16
2x²+22x-21

answer is
2x²+22x-21

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How many permutations of the 26 letters of the English alphabet do not contain any of the strings fish, rat, or bird

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The number of permutations of the 26 letters of the English alphabet that do not contain any of the strings fish, rat, or bird is 402619359782336797900800000

Let

$\mathcal{E}=\{\text{All lowercase letters of the English Alphabet}\}\\\\B=\overline{\{b,i,r,d\}} \cup \{bird\}\\\\F=\overline{\{f,i,s,h\}} \cup \{fish\}\\\\R=\overline{\{r,a,t\}} \cup \{rat\}\\\\FR=\overline{\{f,i,s,h,r,a,t\}} \cup \{fish,rat\}$

Then

$Perm(\mathcal{E})=\{\text{All orderings of all the elements of } \mathcal{E}\}\\\\Perm(B)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing bird}\}\\\\Perm(F)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing fish}\}\\\\Perm(R)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing rat}\}\\\\Perm(FR)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing both fish and rat}\}\\$

Note that since

$F \cap R=\varnothing$ , $Perm(F)\cap Perm(R)\ne \varnothing$

But since

$B \cap R \ne \varnothing$ , $Perm(B)\cap Perm(R)= \varnothing$

and

$B \cap F \ne \varnothing$ , $Perm(B)\cap Perm(F)= \varnothing$

Since

$|\mathcal{E} |=26 \text{, then, } |Perm(\mathcal{E})|=26! \\\\|B|=26-4+1=23 \text{, then, } |Perm(B)|=23!\\\\|F|=26-4+1=23 \text{, then, } |Perm(F)|=23!\\\\|R|=26-3+1=24 \text{, then, } |Perm(R)|=24!\\\\|FR|=26-7+2=21 \text{, then, } |Perm(FR)|=21!\\$