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Ksenya-84 [330]

3 years ago

11

Find the first, fourth, and eighth terms of the sequence. A(n) = –5 • 2n–1

1 answer:

GenaCL600 [577]3 years ago

8 0

Answer:

-5; -40; -640

Step-by-step explanation:

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Anlatarak cevaplarsanız sevinirim

IgorLugansk [536]

Answer:

A)26

Step-by-step explanation:

12 nin dogal sayi bolenleri 1,2,3,4,6 ve 12 dir yani ucgen 6 dir

20 nin en kucuk kati 20 yani kendisidir(20 kendisinin 1 katidir)

5 0

3 years ago

(3x+2)∧2=9 how do i solve this in the square root property

Irina18 [472]

$(3x+2)^2=9 \\\\9x^2+12x+4-9=0\\\\9x^2+12x-5=0\\\\a=9,\ \ b=12, \ \ c=-5$

$x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-12-\sqrt{12^2-4 \cdot9 \cdot (-5)}}{2 \cdot 9}=\frac{-12-\sqrt{144+180}}{18}=\\\\=\frac{-12-\sqrt{324}}{18}=\frac{-12-18}{18}=\frac{-30}{18}=-\frac{5}{3}\\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-12+\sqrt{12^2-4 \cdot9 \cdot (-5)}}{2 \cdot 9}=\frac{-12+18}{18}=\frac{6}{18}=\frac{1}{3} \\\\Answer: \ x=-\frac{5}{3}\ \ and \ \ x=\frac{1}{3}$

6 0

3 years ago

Write an equation for the line parallel to the line 5/8x−7/6y =2 through the point (7,−1).

LiRa [457]

$y = - \frac{28}{48x - 15} - \frac{125}{321}$ is the answer.

Step-by-step explanation:

For this question,

•<u> </u><u>First</u>, find the slope of the line parallel to the line 5/8x−7/6y=2 :-

$m = - \frac{28}{ - 15 + 48x}$

•<u>Second</u>, find the equation of the line,

$y = - \frac{28}{48x - 15} - \frac{125}{321}$

7 0

2 years ago

Solve the equation?

stiks02 [169]

Answer:

A

Step-by-step explanation:

Given

p² + 2p - 8 = 0 ← in standard form

with a = 1, b = 2, c = - 8

Using the quadratic formula to solve for p

p = ( - 2 ± $\sqrt{2^2-(4(1)(-8))}$ / 2

= ( - 2 ± $\sqrt{4+32}$ / 2

= ( - 2 ± $\sqrt{36}$ ) / 2

p = $\frac{-2-6}{2}$ = $\frac{-8}{2}$ = - 4

p = $\frac{-2+6}{2}$ = $\frac{4}{2}$ = 2

5 0

3 years ago

Read 2 more answers

A student solves the following equation and

Phoenix [80]

Answer:

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

Step-by-step explanation:

Considering the expression

$\frac{3}{a+2}-6\cdot \frac{a}{-4+a^2}=\frac{1}{a-2}$

$\frac{3}{a+2}-\frac{6a}{-4+a^2}=\frac{1}{a-2}$

$\mathrm{Find\:Least\:Common\:Multiplier\:of\:}a+2,\:-4+a^2,\:a-2:\quad \left(a+2\right)\left(a-2\right)$

$\mathrm{Multiply\:by\:LCM=}\left(a+2\right)\left(a-2\right)$

$\frac{3}{a+2}\left(a+2\right)\left(a-2\right)-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right)=\frac{1}{a-2}\left(a+2\right)\left(a-2\right)$

as

$\frac{3}{a+2}\left(a+2\right)\left(a-2\right):\quad 3\left(a-2\right)$

$-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right):\quad -6a$

$\frac{1}{a-2}\left(a+2\right)\left(a-2\right):\quad a+2$

so equation becomes

$3\left(a-2\right)-6a=a+2$

$-3a-6=a+2$

$-3a-6+6=a+2+6$

$-4a=8$

$\mathrm{Divide\:both\:sides\:by\:}-4$

$\frac{-4a}{-4}=\frac{8}{-4}$

$a=-2$

$\mathrm{Verify\:Solutions}$

$\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{3}{a+2}-6\frac{a}{-4+a^2}-\frac{1}{a-2}\mathrm{\:and\:compare\:to\:zero}$

$\mathrm{Solve\:}\:a+2=0:\quad a=-2$

$\mathrm{Solve\:}\:-4+a^2=0:\quad a=2,\:a=-2$

$\mathrm{Solve\:}\:a-2=0:\quad a=2$

So the following points are undefined

$a=-2,\:a=2$

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

4 0

3 years ago