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Vika [28.1K]
3 years ago
14

Find each product(6v^2+2v+7)(2v^2-2v+6)​

Mathematics
2 answers:
elixir [45]3 years ago
8 0

Answer:

12v^4-8v^3+46v^2-2v+46

Step-by-step explanation:

remove parentheses and collect like terms

Montano1993 [528]3 years ago
5 0
12v^4-8v^3+46v^2-2v+46

remove parentheses and collect like terms
You might be interested in
Find the equation of the line (in slope-intercept form) through the points (1,-4) and with slope 5/2.
Phoenix [80]

Answer:

Equation of a line is y = mx + c

Where m is the slope

c is the y intercept

Equation of the line using point

( 1 , - 4) and slope 5/2 is

y + 4 =  \frac{5}{2} (x - 1) \\  \\ y + 4 =  \frac{5}{2} x -  \frac{5}{2}  \\  \\ y =  \frac{5}{2} x -  \frac{5}{2}  - 4 \\  \\  \\ y =  \frac{5}{2} x -  \frac{13}{2}

Hope this helps you

8 0
4 years ago
8 + z = 19. step equations
steposvetlana [31]

Answer:

11

Step-by-step explanation:

First subtract

19-8= Z

So solution would be

Z=11

7 0
3 years ago
Read 2 more answers
VEEL
Andre45 [30]

Answer:

a_n=-3(3)^{n-1} ; {-3,-9, -27,- 81, -243, ...}

a_n=-3(-3)^{n-1} ; {-3, 9,-27, 81, -243, ...}

a_n=3(\frac{1}{2})^{n-1} ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

a_n=243(\frac{1}{3})^{n-1} ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

a_n=-3(3)^{n-1}

At n=1,

a_1=-3(3)^{1-1}=-3

At n=2,

a_2=-3(3)^{2-1}=-9

At n=3,

a_3=-3(3)^{3-1}=-27

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

a_n=-3(-3)^{n-1}

At n=1,

a_1=-3(-3)^{1-1}=-3

At n=2,

a_2=-3(-3)^{2-1}=9

At n=3,

a_3=-3(-3)^{3-1}=-27

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

a_n=3(\frac{1}{2})^{n-1}

At n=1,

a_1=3(\frac{1}{2})^{1-1}=3

At n=2,

a_2=3(\frac{1}{2})^{2-1}=1.5

At n=3,

a_3=3(\frac{1}{2})^{3-1}=0.75

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

a_n=243(\frac{1}{3})^{n-1}

At n=1,

a_1=243(\frac{1}{3})^{1-1}=243

At n=2,

a_2=243(\frac{1}{3})^{2-1}=81

At n=3,

a_3=243(\frac{1}{3})^{3-1}=27

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0
4 years ago
Plzzz help me please I need the answers like right now so I’m counting on u guys to help me please who ever is correct get the B
postnew [5]

Sorry but you are cheating yourself but cheating in exams

I know but I will not say you real understand by yourself cheating is big than dying

5 0
3 years ago
Read 2 more answers
Need some help with this please
saveliy_v [14]
-0.83 for it’s the lowest number
5 0
3 years ago
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