Factor each expression completely. x^2 - 25

2 answers:

6 0

We can write it in the form of

x^²-5^²

we know that it's a form of (a^²-b^²)

(a+b) (a-b)

so, (X+5) (x-5)

hope it help you

Nady [450]4 years ago

4 0

Rewrite it in the form a^2 - b^2, where a = x and b = 5

x^2 - 5^2

Use the Difference of Squares: a^2 - b^2 = (a + b)(a - b)

= (x + 5)(x - 5)

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A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f

Andrews [41]

Answer:

$A = \frac{P}{r}\left( e^{rt} -1 \right)$

Step-by-step explanation:

This is a separable differential equation. Rearranging terms in the equation gives

$\frac{dA}{rA+P} = dt$

Integration on both sides gives

$\int \frac{dA}{rA+P} = \int dt$

where $c$ is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

$\int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c$