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identity - an equality that holds true regardless of the values chosen for its variables

1 multiplied by b squared multiplied by c squared

1 multiplied by b squared multiplied by c squared
b squared means we raise b to the power of 2:
b^2
c squared means we raise c to the power of 2:
c^2
b squared multiplied by c squared
b^2c^2
1 multiplied by b squared multiplied by c squared means we multiply 1 by b^2c^2
1b^2c^2
Multiplying by 1 can be written by [U][I]removing[/I][/U] the 1 since it's an identity multiplication:
[B]b^2c^2[/B]

3^14/27^4 = ?

3^14/27^4 = ?
Understand that 27 = 3^3. Rewriting this, we have:
3^14/(3^3)^4
Exponent identity states (a^b)^c = a^bc, so we have:
3^14/3^12
Simplifying, we have:
3^(14 - 12) = 3^2 = [B]9[/B]
[MEDIA=youtube]u_238Z7Mqk[/MEDIA]

A super deadly strain of bacteria is causing the zombie population to double every day. Currently, t

A super deadly strain of bacteria is causing the zombie population to double every day. Currently, there are 25 zombies. After how many days will there be over 25,000 zombies?
We set up our exponential function where n is the number of days after today:
Z(n) = 25 * 2^n
We want to know n where Z(n) = 25,000.
25 * 2^n = 25,000
Divide each side of the equation by 25, to isolate 2^n:
25 * 2^n / 25 = 25,000 / 25
The 25's cancel on the left side, so we have:
2^n = 1,000
Take the natural log of each side to isolate n:
Ln(2^n) = Ln(1000)
There exists a logarithmic identity which states: Ln(a^n) = n * Ln(a). In this case, a = 2, so we have:
n * Ln(2) = Ln(1,000)
0.69315n = 6.9077
[URL='https://www.mathcelebrity.com/1unk.php?num=0.69315n%3D6.9077&pl=Solve']Type this equation into our search engine[/URL], we get:
[B]n = 9.9657 days ~ 10 days[/B]

Additive Identity Property

Free Additive Identity Property Calculator - Displays the line by line proof for the additive identity property
Numerical Properties

Approximate Square Root Using Exponential Identity

Free Approximate Square Root Using Exponential Identity Calculator - Calculates the square root of a positive integer using the Exponential Identity Method

cscx-cotx*cosx=sinx

cscx-cotx*cosx=sinx
A few transformations we can make based on trig identities:
[LIST]
[*]csc(x) = 1/sin(x)
[*]cot(x) = cos(x)/sin(x)
[/LIST]
So we have:
1/sin(x) - cos(x)/sin(x) * cos(x) = sin(x)
(1 - cos^2(x))/sin(x) = sin(x)
1 - cos^2(x) = sin^2(x)
This is [B]true[/B] from the identity:
sin^2(x) - cos^2(x) = 1

Euclids Algorithm and Euclids Extended Algorithm

Free Euclids Algorithm and Euclids Extended Algorithm Calculator - Given 2 numbers a and b, this calculates the following

1) The Greatest Common Divisor (GCD) using Euclids Algorithm

2) x and y in Bézouts Identity ax + by = d using Euclids Extended Algorithm Extended Euclidean Algorithm

1) The Greatest Common Divisor (GCD) using Euclids Algorithm

2) x and y in Bézouts Identity ax + by = d using Euclids Extended Algorithm Extended Euclidean Algorithm

Express cos4θ and sin4θ in terms of sines and cosines of multiples of θ

Express cos4θ and sin4θ in terms of sines and cosines of multiples of θ.
Using a trignometric identity:
cos (2θ) = cos^2(θ) - sin^2(θ)
Since 4θ = 2*2θ, so we have:
[B]cos(4θ) = cos^2(2θ) - sin^2(2θ)[/B]
Using another trignometric identity, we have:
sin(2θ) = 2 sin(θ) cos(θ)
Since 4θ = 2*2θ, so we have:
[B]sin(4θ) = 2 sin(2θ) cos(2θ)[/B]

How can you rewrite the number 1 as 2 to a power?

How can you rewrite the number 1 as 2 to a power?
There exists an identity which says, n^0 = 1 where n is a number.
So [B]2^0 = 1[/B]

if Logb(5)=3.56 and logb(8)=4.61 then what is logb(40)

if Logb(5)=3.56 and logb(8)=4.61 then what is logb(40)
There exists a logarithmic identity which says log(xy) = log(x) + log(y).
Since the two logs above have the same base b, we have:
x = 5 and y = 8. So we have:
logb(40) = logb(5) + logb(8)
logb(40) = 3.56 + 4.61
logb(40) = [B]8.17[/B]

log5 = 0.699, log2 = 0.301. Use these values to evaluate log40

log5 = 0.699, log2 = 0.301. Use these values to evaluate log40.
One of the logarithmic identities is: log(ab) = log(a) + log(b). Using the numbers 2 and 5, we somehow need to get to 40.
[URL='http://www.mathcelebrity.com/factoriz.php?num=40&pl=Show+Factorization']List factors of 40[/URL].
On the link above, take a look at the bottom where it says prime factorization. We have:
40 = 2 x 2 x 2 x 5
Using our logarithmic identity, we have:
log40 = log(2 x 2 x 2 x 5)
Rewriting this using our identity, we have:
log40 = log2 + log2 + log2 + log5
log40 = 0.301 + 0.301 + 0.301 + 0.699
log40 = [B]1.602
[MEDIA=youtube]qyG_Jkf9VDc[/MEDIA][/B]

Multiplicative Identity Property

Free Multiplicative Identity Property Calculator - Demonstrates the Multiplicative Identity property using a number.
Numerical Properties

Oliver invests $1,000 at a fixed rate of 7% compounded monthly, when will his account reach $10,000?

Oliver invests $1,000 at a fixed rate of 7% compounded monthly, when will his account reach $10,000?
7% monthly is:
0.07/12 = .00583
So we have:
1000(1 + .00583)^m = 10000
divide each side by 1000;
(1.00583)^m = 10
Take the natural log of both sides;
LN (1.00583)^m = LN(10)
Use the identity for natural logs and exponents:
m * LN (1.00583) = 2.30258509299
0.00252458479m = 2.30258509299
m = 912.064867899
Round up to [B]913 months[/B]