I cant do it sombody help

Mathematics

1 answer:

motikmotik3 years ago

3 0

Step-by-step explanation:

hope this helps. in the first part you set the problems equal to each other. in the second, you get rid of one letter, then solve.

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Find the 20th term in the sequence, where the first term is -2 and the common ratio is-2.

geniusboy [140]

Answer:

Step-by-step explanation:

(1)

a1=-2

r=-2

$a_{20}=(-2)(-2)^{20-1}=-2(-2)^{19}=2^{20}=1048576\\(2)\\a_{1}=1000\\r=\frac{200}{1000} =\frac{1}{5} \\a_{8}=1000(\frac{1}{5} )^{8-1}=1000(.2)^7=0.0128\\(3)\\a_{1}=2\\a_{3}=50\\a_{2}=\sqrt{2*50} =10\\r=\frac{10}{2} =5$

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Find the product of (2x + 5y)(2x − 5y).

Nana76 [90]

The final answer for this is 4x^2-25y^2.

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Read 2 more answers

Solve the equation: k^2+5k+13=0

mr_godi [17]

Step-by-step explanation:

k² + 5k + 13 = 0

Using the quadratic formula which is

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\$

From the question

a = 1 , b = 5 , c = 13

So we have

$k = \frac{ - 5 \pm \sqrt{ {5}^{2} - 4(1)(13) } }{2(1)} \\ = \frac{ - 5 \pm \sqrt{25 - 52} }{2} \\ = \frac{ - 5 \pm \sqrt{ - 27} }{2} \: \: \: \: \: \: \\ = \frac{ - 5 \pm3 \sqrt{3} \: i}{2} \: \: \: \: \: \:$