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tensa zangetsu [6.8K]
3 years ago
11

Which of the following is the slope of the line that passes through the points ( –2, 8) and (10, –1)

Mathematics
1 answer:
prohojiy [21]3 years ago
8 0

Answer:

none of them

Step-by-step explanation:

plug it into desmos, and none of them pass through the points

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Solve 4 + 4 ( x - 2 ) = 2 ( x + 1 ) - x<br> The question requires you to show your work
kati45 [8]

Answer:

x = 2

Step-by-step explanation:

Step  1  :

Equation at the end of step  1  :

 (4+(4•(x-2)))-(2•(x+1)-x)  = 0

Step  2  :

Equation at the end of step  2  :

 (4 +  4 • (x - 2)) -  (x + 2)  = 0

Step  3  :

Step  4  :

Pulling out like terms :

4.1     Pull out like factors :

  3x - 6  =   3 • (x - 2)

Equation at the end of step  4  :

 3 • (x - 2)  = 0

Step  5  :

Equations which are never true :

5.1      Solve :    3   =  0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

5.2      Solve  :    x-2 = 0

Add  2  to both sides of the equation :

                     x = 2

One solution was found :

                  x = 2

Processing ends successfully

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4 years ago
What is the area of this figure?<br> Enter your answer in the box.  __in²
dexar [7]
Just add up the component areas- the 2 triangles, and then break the remainder into a rectangle and a triangle.

4 0
4 years ago
22. An employee joined a company in 2017 with a starting salary of $50,000. Every year this employee receives a raise of $1000 p
Setler79 [48]

Answer:

(a) The recurrence relation for the salary is

S_{n+1}=1.05*S_n+1000\\\\S_0=50000

(b) The salary 25 years after 2017 will be $217044.85.

(c) S_n=1.05^nS_0+1000*\sum_{0}^{n-1}1.05^n

Step-by-step explanation:

We can define the next year salary S_{n+1} as

S_{n+1}=S_n+1000+0.05*S_n=1.05*S_n+1000

wit S0=$50000

If we extend this to 2 years from 2017 (n+2), we have

S_{n+2}=1.05*S_{n+1}+1000=1.05*(1.05*S_n+1000)+1000\\S_{n+2} =1.05^2*S_n+1.05*1000+1000\\S_{n+2}=1.05^2*S_n+1000*(1.05^1+1)

Extending to 3 years (n+3)

S_{n+3}=1.05*S_{n+2}+1000=1.05(1.05^2*S_n+1000*(1.05^1+1))+1000\\\\S_{n+3}=1.05^3S_n+1.05*1000*(1.05^1+1)+1000\\\\S_{n+3}=1.05^3*S_n+1000*(1.05^2+1.05^1+1)

Extending to 4 years (n+4)

S_{n+4}=1.05*S_{n+3}+1000=1.05*(1.05^3*S_n+1000*(1.05^2+1.05^1+1))+1000\\\\S_{n+4}=1.05^4S_n+1.05*1000*(1.05^2+1.05^1+1))+1000\\\\S_{n+4}=1.05^4S_n+1000*(1.05^3+1.05^2+1.05^1+1.05^0)

We can now express a general equation for S_n (salary at n years from 2017)

S_n=1.05^nS_0+1000*\sum_{0}^{n-1}1.05^n

The salary at 25 years from 2017 (n=25) will be

S_{25}=1.05^{25}S_0+1000*\sum_{0}^{24}1.05^i\\\\S_{25}=3.386*50000+1000*47.72=217044.85

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Answer: 48x-12

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