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3 years ago

Find the 15th term of the geometric sequence 3, -6, 12, ...

Mathematics

1 answer:

Roman55 [17]3 years ago

4 0

Answer:

49152

Step-by-step explanation:

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Help with algebra 2 I can’t figure out how to solve the problem .

defon

Answer: D

Step-by-step explanation:

Using the equation plot in the coordinates of the points to see if they fit the equation.

5x-3y < 30

A) 5(-3) -3(5) < 30

-15 -15 < 30

-30 < 30 This is true

B) 5(0) -3(0) < 30

0 - 0 < 30

0< 30 This is true

C) 5(-5) - 3(3) < 30

-25 - 9 < 30

-34 < 30 This is also true

D) 5(3) - 3(-5) < 30

15 + 15 < 30

30 < 30 This is not true because 30 is not greater that 30

3 0

3 years ago

How do you find the angle measure?

yulyashka [42]

Answer:

∠ XYZ = 41°

Step-by-step explanation:

∠ XYZ = ∠ 3x + 20 ( corresponding angles ) , thus

8x - 15 = 3x + 20 ( subtract 3x from both sides )

5x - 15 = 20 ( add 15 to both sides )

5x = 35 ( divide both sides by 5 )

x = 7

Thus

∠ XYZ = 8x - 15 = 8(7) - 15 = 56 - 15 = 41°

5 0

3 years ago

Suppose that contamination particle size (in micrometers) canbe modeled as

asambeis [7]

Answer:

c) 2

d) 0.96

Step-by-step explanation:

We are given the following in the question:

$f(x) = 2x^{-3}, x > 1\\~~~~~~~= 0, x \leq 1$

a) probability density function.

$\displaystyle\int^{\infty}_{\infty}f(x) dx = 1\\\\\displaystyle\int^{\infty}_{-\infty}2x^{-3}dx = 1\\\\\displaystyle\int^{\infty}_{1}2x^{-3}dx\\\\\Rightarrow \big[-x^{-2}\big]^{\infty}_1\\\\\Rightarrow -(0-1) = 1$

Thus, it is a probability density function.

b) cumulative distribution function.

$P(X$

c) mean of the distribution

$\mu = \displaystyle\int^{\infty}_{-\infty}xf(x) dx\\\\\mu = \int^{\infty}_12x^{-2}dx\\\\\mu = (-\frac{2}{x})^{\infty}_{1}\\\\\mu = 2$

d) probability that the size of random particle will be less than 5 micrometers

$P(X$

4 0

3 years ago

Please help, I have been trying for a while :(<br><br>question on photo

Zigmanuir [339]

Answer:

$\frac{x-22}{(x+2)(x-4)}$

Step-by-step explanation:

multiply the numerator/denominator of the first fraction by (x - 4)

multiply the numerator/denominator of the second fraction by (x + 2)

this ensures that the fractions have a common denominator

$\frac{4}{x+2}$ - $\frac{3}{x-4}$

= $\frac{4(x-4)}{(x+2)(x-4)}$ - $\frac{3(x+2)}{(x+2)(x-4)}$ ← subtract numerators leaving the common denominator

= $\frac{4x-16-3x-6}{(x+2)(x-4)}$

= $\frac{x-22}{(x+2)(x-4)}$

4 0

2 years ago

Please help quickly as possible thank you!

tensa zangetsu [6.8K]

Multiply $Q\cdot P$ , quantity and price,

$1.2\cdot5.95=7.14$ .

Hope this helps :)

5 0

3 years ago

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