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Tamiku [17]
3 years ago
5

Write three more expressions using exponents that evaluate to 216.

Mathematics
1 answer:
kow [346]3 years ago
7 0
6^{3}
216^{1}
46656^{0.5}
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8090 [49]
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6 0
3 years ago
What is the distance between (-6,4) and (-8, 6)?
Tatiana [17]

Answer:

\sqrt{8}=2\sqrt{2}

Step-by-step explanation:

by using well known formula, distance on XOY plane

\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\=\sqrt{(-8-(-6))^2+(6-4)^2}\\=\sqrt{(-2)^2+(2)^2}

8 0
3 years ago
23. Tell whether the sequence is arithmetic. If it is, what is the common difference? (1
Airida [17]

Answer:   O Yes; 6

Yes it is Arithmetic;  the difference is 6

Step-by-step explanation:

If the numbers keep changing by the same value being added or subtracted, it is Arithmetic.

-15 - (-9) = -6    -9 - (-3) = -6    -3 - 3 = -6

"Same difference"  as the saying goes!

4 0
3 years ago
A point is randomly chosen on a map of North America. Describe the probability of the point being in each location: North Americ
Aleks04 [339]

Answer:

We know that the map is of North America:

The probabilities are:

1) North America:

As the map is a map of North America, you can point at any part of the map and you will be pointing at North America, so the probability is p = 1

or 100% in percentage form.

2) New York City.

Here we can think this as:

The map of North America is an extension of area, and New Yorck City has a given area.

As larger is the area of the city, more probable to being randomly choosen, so to find the exact probability we need to find the quotient between the area of New York City and the total area of North America:

New York City = 730km^2

North America = 24,709,000 km^2

So the probability of randomly pointing at New York City is:

P = ( 730km^2)/(24,709,000 km^2) = 3x10^-5 or 0.003%

3) Europe:

As this is a map of Noth America, you can not randomly point at Europe in it (Europe is other continent).

So the probaility is 0 or 0%.

5 0
3 years ago
What is the expansion of (3+x)^4
Vlad1618 [11]

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

\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3

\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
6 0
3 years ago
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