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serious [3.7K]
3 years ago
7

Gavin loves biking. He wants to set a new score by riding 65 miles in four days. Each day he wants to ride 1.5 times as far as t

he day before. X is how much he rode the first day, tomorrow he will ride X times 1.5 miles. How far should he ride each day of four days to reach his goal? Write an expression representing the total distance he should bike over four days, do not combine alike terms.
Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
7 0

Answer:How far should he ride on each of the four days to reach his goal?

1st day:   8 miles

2nd day: 12 miles

3rd day:  18 miles

4th day:  27b miles

Step-by-step explanation

u  should vote me branliest hehe

kap26 [50]3 years ago
6 0
1st 8 miles 2nd 12 miles 3rd 18 miles 4th 27b miles
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Answer:

\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81

Step-by-step explanation:

Considering the expression

\left(3+x\right)^4

Lets determine the expansion of the expression

\left(3+x\right)^4

\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i

a=3,\:\:b=x

=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i

Expanding summation

\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}

i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0

i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1

i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2

i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3

i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4

=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4

=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4

as

\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81

\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x

\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2

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\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4

so equation becomes

=81+108x+54x^2+12x^3+x^4

=x^4+12x^3+54x^2+108x+81

Therefore,

  • \left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81
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