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3 years ago

How do you solve Log (base 8) 2 =x

1 answer:

3 0

Log8 2 = x
by the definition of logs, we can write

8^x = 2

8^(1/3) = 2

so x = 1/3

or we could solve it by using log to base 10 and use our calculator

log 8^x = log 2
x log 8 = log 2

x = log 2 / log 8 = 0.333.... = 1/3

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What second degree polynomial function has a leading coefficient of -2 and root 4 with a multiplicity of 2

OverLord2011 [107]

The second degree polynomial with leading coefficient of -2 and root 4 with multiplicity of 2 is:

$p(x) = -2*(x - 4)*(x - 4) = -2*(x - 4)^2$

<h3>How to write the polynomial?</h3>

A polynomial of degree N, with the N roots {x₁, ..., xₙ} and a leading coefficient a is written as:

$p(x) = a*(x - x_1)*(x - x_2)*...*(x - x_n)$

Here we know that the degree is 2, the only root is 4 (with a multiplicity of 2, this is equivalent to say that we have two roots at x = 4) and a leading coefficient equal to -2.

Then this polynomial is equal to:

$p(x) = -2*(x - 4)*(x - 4) = -2*(x - 4)^2$

If you want to learn more about polynomials:

brainly.com/question/4142886

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8 0

2 years ago

Which scale will produce the smallest drawing?

Effectus [21]

The scale that will produce the smallest drawing would be b) 1mm:50m.

First, I converted all the measurements to centimeters:

a) 1cm:500cm
b) 0.01cm:5000cm
c) 5cm:1000cm
d) 10cm:2500cm

Then, you divide the scales:

a) 0.002
b) 0.000002
c) 0.005
d) 0.004

B is the smallest number, meaning it will be the smallest scale.

6 0

3 years ago

What is the bigger number, a googol or a billion?

Doss [256]

A googol is bigger than a billion.

3 0

4 years ago

Read 2 more answers

Prove algebraically that the recurring decimal 0.72 =8/11

miskamm [114]

Answer:

see explanation

Step-by-step explanation:

We require 2 equations with the recurring decimal placed after the decimal point.

let x = 0.7272.... (1) ← multiply both sides by 100

100x = 72.7272... (2)

Subtract (1) from (2) thus eliminating the recurring decimal

99x = 72 ( divide both sides by 99 )

x = $\frac{72}{99}$ = $\frac{8}{11}$

7 0

3 years ago

Read 2 more answers

For the function below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subin

Vlad [161]

Answer:

See below

Step-by-step explanation:

We start by dividing the interval [0,4] into n sub-intervals of length 4/n

$[0,\displaystyle\frac{4}{n}],[\displaystyle\frac{4}{n},\displaystyle\frac{2*4}{n}],[\displaystyle\frac{2*4}{n},\displaystyle\frac{3*4}{n}],...,[\displaystyle\frac{(n-1)*4}{n},4]$

Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.

Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)

$\displaystyle\frac{4}{n}f(\displaystyle\frac{1*4}{n})+\displaystyle\frac{4}{n}f(\displaystyle\frac{2*4}{n})+...+\displaystyle\frac{4}{n}f(\displaystyle\frac{n*4}{n})=\\\\=\displaystyle\frac{4}{n}((\displaystyle\frac{1*4}{n})^2+3+(\displaystyle\frac{2*4}{n})^2+3+...+(\displaystyle\frac{n*4}{n})^2+3)=\\\\\displaystyle\frac{4}{n}((1^2+2^2+...+n^2)\displaystyle\frac{4^2}{n^2}+3n)=\\\\\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12$

but

$1^2+2^2+...+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}$

so the upper sum equals

$\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12=\displaystyle\frac{4^3}{n^3}\displaystyle\frac{n(n+1)(2n+1)}{6}+12=\\\\\displaystyle\frac{4^3}{6}(2+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2})+12$

When $n\rightarrow \infty$ both $\displaystyle\frac{3}{n}$ and $\displaystyle\frac{1}{n^2}$ tend to zero and the upper sum tends to

$\displaystyle\frac{4^3}{3}+12=\displaystyle\frac{100}{3}$

8 0

4 years ago