How to solve 10(x+4)(x-4)

Mathematics

1 answer:

IceJOKER [234]3 years ago

4 0

Answer:

1 Use Difference of Squares:

a^2-b^2=(a+b)(a−b).

10(x^2-4^2)

2 Simplify 4^2 to 16

10(x^2 −16)

3 Expand by distributing terms.

10x^2 −160

Step-by-step explanation:

Hope dis helps and pls mark me as brainlist!!!

§ALEX§

You might be interested in

What measurement is equal to 24 kilograms

Vladimir79 [104]

52.9109 in pounds if that's what your asking.

3 0

4 years ago

Read 2 more answers

Find the rate of change

Tatiana [17]

$\bf slope = m = \cfrac{rise}{run} \implies 
\cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby 
\begin{array}{llll}
average~rate\\
of~change
\end{array}\\\\
-------------------------------\\\\
h(x)= 2^x \qquad 
\stackrel{2\le x\le 4}{\begin{cases}
x_1=2\\
x_2=4
\end{cases}}\implies \cfrac{h(4)-h(2)}{4-2}\implies \cfrac{2^4-2^2}{2}
\\\\\\
\cfrac{16-4}{2}\implies \cfrac{12}{2}\implies 6$

5 0

3 years ago

How do you this question (pre-cal)

Novay_Z [31]

The goal to proving identities is to transform one side into the other. We can only pick one side to transform while the other side stays the same the entire time. The general rule of thumb is to transform the more complicated side (though there may be exceptions to this guideline).

So I'll take the left hand side and try to turn it into $\csc^2( B )$

One way we can do that is through the following steps:

$\frac{\tan(B) + \cot(B)}{\tan(B)} = \csc^2(B)\\\\\frac{\tan(B)}{\tan(B)} + \frac{\cot(B)}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\frac{1}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\cot(B) = \csc^2(B)\\\\1 + \cot^2(B) = \csc^2(B)\\\\1 + \frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)}{\sin^2(B)}+\frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)+cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{1}{\sin^2(B)} = \csc^2(B)\\\\\csc^2(B)=\csc^2(B) \ \ {\Large \checkmark}\\\\$