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Mathematics

1 answer:

ss7ja [257]2 years ago

5 0

Line t. It cuts directly into the angle. Don’t get it confused with bisect because that comes directly from the angle and not just anywhere on it. Hope this helps

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Whats the answer to 8p squared minus 16 p=10

goblinko [34]

8p² - 16p = 10

8p² - 16p - 10 = 0 Divide through by 2

4p² - 8p - 5 = 0

Multiply first and last coefficients: 4*-5 = -20

We look for two numbers that multiply to give -20, and add to give -8

Those two numbers are 2 and -10.

Check: 2*-10 = -20 2 + -10 = -8

We replace the middle term of -8p in the quadratic expression with 2p -10p

4p² - 8p - 5 = 0

4p² + 2p - 10p - 5 = 0

2p(2p + 1) - 5(2p + 1) = 0

(2p + 1)(2p - 5) = 0

2p + 1 = 0 or 2p + 5 = 0

2p = 0 -1 2p = 0 - 5

2p = -1 2p = -5

p = -1/2 p = -5/2

The solutions are p = -1/2 or -5/2

7 0

3 years ago

Read 2 more answers

What is the 7th term of the geometric sequence where a1 = -625 and a2 = 125

kykrilka [37]

Since the sequence is geometric, there is some constant $r$ such that the sequence is recursively given by

$a_n=ra_{n-1}$

By this definition, you can recursively substitute into the right hand side the definition for $a_{n-1},a_{n-2},\ldots$ to find an explicit formula for the $n$ th term.

$a_n=ra_{n-1}=r^2a_{n-2}=r^3a_{n-3}=\cdots=r^{n-1}a_1=-625r^{n-1}$

You know the second term, which means you can find $r$ :

$a_2=-625r^{2-1}\implies125=-625r\implies r=-\dfrac15$

So, the 7th term of the sequence is

$a_7=-625\left(-\dfrac15\right)^{7-1}=-\dfrac1{25}$