So distribute using distributive property
a(b+c)=ab+ac so
split it up
(5x^2+4x-4)(4x^3-2x+6)=(5x^2)(4x^3-2x+6)+(4x)(4x^3-2x+6)+(-4)(4x^3-2x+6)=[(5x^2)(4x^3)+(5x^2)(-2x)+(5x^2)(6)]+[(4x)(4x^3)+(4x)(-2x)+(4x)(6)]+[(-4)(4x^3)+(-4)(-2x)+(-4)(6)]=(20x^5)+(-10x^3)+(30x^2)+(16x^4)+(-8x^2)+(24x)+(-16x^3)+(8x)+(-24)
group like terms
[20x^5]+[16x^4]+[-10x^3-16x^3]+[30x^2-8x^2]+[24x+8x]+[-24]=20x^5+16x^4-26x^3+22x^2+32x-24
the asnwer is 20x^5+16x^4-26x^3+22x^2+32x-24
Answer:
174
Step-by-step explanation:
L = 68
W=19
Use the formula given:
2 x 68 + 2 x 19 = 174
A=LW
192=(13+x)((9+x)
192=117+22x+x^2 (subtract 192 from both sides)
x^2+22x-75+0 (factor)
(x+25)(x-3)=0
so x=-25 or 3
Check work
since we are adding x it can only be the positive solution so x=3 so the new dimensions are:
L=13+3=16m
W=9+3=12
check: 16*12=192 so the answer is correct hope this helped
Answer: 
.
Step-by-step explanation:
Formula : Probability = 
From the given frequency distribution for the class level of students in an introductory statistics course, we have
Number of Junior= 12
Number of Senior = 7
Total students = 6+16+12+7 = 41
Probability that the first student obtained is a junior = 
Probability that the the second student obtained is a senior. 
Then, the probability that the first student obtained is a junior and the second a senior would be 
Hence, the required probability is .
