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3 years ago

Simplify the expression 2^3 x 2^2

Mathematics

2 answers:

SCORPION-xisa [38]3 years ago

8 0

Simply add the exponents to be 2^5.

SpyIntel [72]3 years ago

3 0

It would be 2^5 which equals 32

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Use the substitution method to solve the given system of equations. Write your answer as an ordered pair, (x, y).

Gnoma [55]

Answer:

(-3, -2)

Step-by-step explanation:

Look at the second equation, we can easily rearrange it to make x the subject.

x - 4y = 5

x = 4y + 5

Now lets put that into the second equation

2x - 2y = -2

2(4y + 5) - 2y = -2

6y + 10 = -2

6y = -12

y = -2

Now lets find x!

x = 4y + 5

x = 4(-2) + 5

x = -3

Now lets format the answer like the question says (ordered pair)

(-3, -2)

Recap: We used substitution and made an equation to work out one of the variables first, then found the second variable using the first.

3 0

3 years ago

Read 2 more answers

Multiply the binomial and the trinomial. (x-2)(x^2+2x+4)

malfutka [58]

Answer:

Step-by-step explanation:

To solve this problem, we need to multiple $x$ and -2 by $x^{2} + 2x + 4$ and add the results together:

$x(x^{2} + 2x + 4)$

$x^{3} + 2x^{2} + 4x$

$-2(x^{2} + 2x + 4)$

$-2x^{2} - 4x - 8$

Adding the two together give the following:

$(x^{3} + 2x^{2} + 4x) + (-2x^{2} - 4x - 8)$

$x^{3} - 8$

7 0

3 years ago

The heights of a maple tree and a cherry tree have a ratio of 5:2. If the maple tree grew 20 cm and 20 cm was cut off the top of

elena-14-01-66 [18.8K]

Let x be the heights of a maple tree and y be the height of the cherry tree.
We know:
 $\frac{x}{y} = \frac{5}{2}$
The new ratio is obtained like this:
 $\frac{x+20}{y-20} = \frac{3}{1}$ .
From the above equation we get $x= (y-20)\frac{3}{1} -20$ .
then $\frac{ (y-20)\frac{3}{1} -20}{y} = \frac{5}{2}$
Solving the above equation for y we get: $y=160$
So $x=400$
So the first tree is (400-160=240) more taller than cherry tree.

7 0

3 years ago

A particular geometric sequence has strictly decreasing terms. After the first term, each successive term is calculated by multi

Maksim231197 [3]

Answer:

6 possible integers

Step-by-step explanation:

Given

A decreasing geometric sequence

$Ratio = \frac{m}{7}$

Required

Determine the possible integer values of m

Assume the first term of the sequence to be positive integer x;

The next sequence will be $x * \frac{m}{7}$

The next will be; $x * (\frac{m}{7})^2$

The nth term will be $x * (\frac{m}{7})^{n-1}$

For each of the successive terms to be less than the previous term;

then $\frac{m}{7}$ must be a proper fraction;

This implies that:

$0 < m < 7$

Where 7 is the denominator

The sets of $0 < m < 7$ is $\{1,2,3,4,5,6\}$ and their are 6 items in this set

Hence, there are 6 possible integer

3 0

3 years ago

Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

tekilochka [14]

Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$

Applying the ratio test:

$\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}$

$\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

Applying n → ∞

$\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

The numerator as well denominator of $\frac{a_{n}}{a_{n+1}}$ are polynomials of fifth degree with leading coefficients:

$(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}$

4 0

3 years ago