If his best time is x, and the best time is smaller, then (1+1/3)*x=his second best time. If x+4=his second best time (since his second best time is 4 seconds slower), then (1+1/3)*x=x+4. Multiplying it out, we get (3/3+1/3)*x=x+4
=4x/3=x+4. Subtracting x from both sides, we get 4x/3-3x/3=x/3=4. Multiplying both sides by 3, we get x=12=his best time

8 0

4 years ago

(5x^5) - (80x^3)<br> factoring

elena55 [62]

Answer:

Factoring the term $(5x^5) - (80x^3)$ we get $\mathbf{5x^3(x-4)(x+4)}$

Step-by-step explanation:

We need to factor the term $(5x^5) - (80x^3)$

First we can see that $5x^3$ is common in both terms

So, taking $5x^3$ common:

$(5x^5) - (80x^3)\\=5x^3(x^2-16)$

We can write $x^2-16$ as $(x)^2-(4)^2$

$=5x^3((x)^2-(4)^2)$

Now we can solve $(x)^2-(4)^2$ using the formula: $a^2-b^2=(a+b)(a-b)$

We can write:

$=5x^3((x)^2-(4)^2)\\=5x^3(x-4)(x+4)$

So, factoring the term $(5x^5) - (80x^3)$ we get $\mathbf{5x^3(x-4)(x+4)}$

4 0

3 years ago

Write 2^8 * 8^2 * 4^-4 in the form 2^n

Eva8 [605]

The given expression 2^8 * 8^2 * 4^-4 can be written in the exponential form 2^n as 2^6.

<h3>What are exponential forms?</h3>

The exponential form is a more convenient way to write repetitive multiplication of the same integer by using the base and its exponents.

<u>For example:</u>

If we have a*a*a*a, it can be written in exponential form as:

=a^4

where

a is the base, and
4 is the power.

The power in this format reflects the number of times we multiply the base by itself. The exponent is also known as the index or power.

From the information given:

We can write 2^8 * 8^2 * 4^-4 in form of 2^n as follows:

$\mathbf{= 2^8\times (2^3)^2 \times (2^2)^{-4} }$