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3 years ago

2^5*2^negative 3 simplify

Mathematics

1 answer:

Citrus2011 [14]3 years ago

5 0

Answer:

Step-by-step explanation:

Simplify the following:

2^5/2^3

Hint: | For all exponents, a^n/a^m = a^(n - m). Apply this to 2^5/2^3.

Combine powers. 2^5/2^3 = 2^(5 - 3):

2^(5 - 3)

Hint: | Subtract 3 from 5.

5 - 3 = 2:

2^2

Hint: | Evaluate 2^2.

2^2 = 4:

Answer: 4

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Axis of symmetry f(x) = −4(x − 8)^ 2 + 3

Delvig [45]

Answer:

$x=8$

Step-by-step explanation:

For an equation in the form of $y=a(x-h)^2+k$ , the axis of symmetry is where $x=h$ . In this case, the axis of symmetry is the line $x=8$

7 0

2 years ago

Which expression is equivalent to log w (x^2 -6)^4/ 3 sqrt x^2+8?

evablogger [386]

Answer:

C $4\log_w(x^2-6)-\dfrac{1}{3}\log_w(x^2+8)$

Step-by-step explanation:

First use the property of logarithms

$\log _ab-\log_ac=\log_a\dfrac{b}{c}.$

For the given expression you get

$\log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2+8} }=\log_w(x^2-6)^4-\log_w\sqrt[3]{x^2+8}=\log_w(x^2-6)^4-\log_w(x^2+8)^{\frac{1}{3}}$

Now use property of logarithms

$\log_ab^k=k\log_ab.$

For your simplified expression, you get

$\log_w(x^2-6)^4-\log_w(x^2+8)^{\frac{1}{3}}=4\log_w(x^2-6)-\dfrac{1}{3}\log_w(x^2+8).$

3 0

3 years ago

30 could be am awnser but so could 28 and 35

STatiana [176]

Answer: definitely 35

Step-by-step explanation:

7 0

4 years ago

What is the value of 3n if n=2

Harman [31]

Answer:

Step-by-step explanation:

3n=3(n)

ur replcing n with 2

3(2)=3x2

3x2=6

6 0

3 years ago

Read 2 more answers

* Some amount of billiard balls were arranged in an equilateral triangle. And 5 balls were extra. When the same set of billiard

Agata [3.3K]

Lets say billiard balls are arranged in rows to form an equilateral triangle, then the first row consists of 1 ball, second row consists of 2 balls, and third row consists of 3 balls, and so on. So there must be $n$ balls in the $n^{th}$ row.

So, the total number of balls that forms the equilateral triangle with $n$ rows is:

$1+2+3+4+5+....+n=\frac{n(n+1)}{2}$

Let $x_1$ and $x_2$ be the total number of balls in the first and second arrangements respectively.

Then,

$x_1=\frac{n(n+1)}{2} +5$

It has been said that there were 11 lesser balls in the second arrangement:

$x_2=\frac{1+(n+1)}{2} \times (n+1)-11=(n+1) \times \frac{(n+2)}{2} -11$

Since, $x_1=x_2$

$\frac{(n+1)}{2} \times n+5=\frac{(n+2)}{2} \times(n+1)-11$

multiplying both the sides by 2

$(n+1)\times n+10=(n+2)(n+1)-22$

$n+n^2=n^2+n+2n+2-22-10$

$2n=22+10-2$

$2n=30$

$n=15$

Therefore,

$x_1=\frac{(n+1)}{2}\times n+5=\frac{15+1}{2} \times 15+5=125$

So, there were 125 balls at the set.

4 0

3 years ago

Read 2 more answers

2^5*2^negative 3 simplify​

2^5*2^negative 3 simplify