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3 years ago

How to factor 2c^2-7c+3

Mathematics

2 answers:

Semmy [17]3 years ago

7 0

(c - 3)(2c - 1) is the factored form

Grace [21]3 years ago

7 0

$2c^2-7c+3$

You need 2 numbers that when added together equal the coefficient of c (-7 in this case), and when multiplied equal the product of the constant and coefficient of c^2 (this product is 2*3 = 6).
Two numbers that add to -7 and multiply to 6 are -6 and -1.

The middle term -7c becomes -6c - 1c.
$2c^2-6c-1c+3 \\ 
2c(c-3) -1(c-3) \\ 
(c-3)(2c-1)$

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3 years ago

Find the area of a circle with a circumference of 6.28 unitesz

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Answer:

3.14 square units

Step-by-step explanation:

Circumference of a circle = 2πr

2πr = 6.28

r = 6.28/2π

r = 3.142/π

Area of a circle = πr^2

slot in the value of r

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3 years ago

Write out the first few terms of the series Summation from n equals 0 to infinity (StartFraction 2 Over 3 Superscript n EndFract

anyanavicka [17]

Answer:

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = 15/8$

Step-by-step explanation:

The sum you are trying to understand is this.

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n}$

Remember that in general when you have a geometric series

$\sum\limits_{n = 0}^{\infty} a*r^n$ you have that

$\sum\limits_{n = 0}^{\infty} a*r^n = \frac{a}{1-r}$ and that equality is true as long as $|r| < 1$ .

Therefore here we have

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{3*5} \big)^n = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n$ and $\big|\frac{-1}{15} \big| = \frac{1}{15} < 1$

Therefore we can use the formula and

$\sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n = \frac{2}{1-(-1/15)} = \frac{2}{1+1/15} = 30/16 = 15/8$

5 0

3 years ago