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Alja [10]
2 years ago
7

1. 8x – 4 = 12x + 12

Mathematics
1 answer:
olga_2 [115]2 years ago
3 0
7(x-2) +3x=46
7x-14+3x=46
10x-14=46
10x=60
X=6
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If f(x)= 1/x and g(x) = 3x+2 find (g • g)(-4)
OleMash [197]

Answer:

-4g2

Step-by-step explanation:

5 0
2 years ago
Please help me simplified this
Lera25 [3.4K]

For this case we have that by definition, the perimeter of the quadrilateral shown is given by the sum of its sides:

Let "p" be the perimeter of the quadrilateral, then:

p = 7 + y + 7 + x\\p = 7 + 7 + x + y\\p = 14 + x + y

So, the perimeter of the figure is: 14 + x + y

Answer:

p = 14 + x + y

7 0
3 years ago
Read 2 more answers
What is the solution to the following system?<br> x+y+z=6 <br> x-y+z=8 <br> x+y-z=0
Ghella [55]
I don't really know if this is correct but I think it is x+y+z=14
4 0
3 years ago
Read 2 more answers
Can I get the answers for number 14 plz?
nataly862011 [7]
A.
-1: 
2(2)^{-1}  \\ 2( \frac{1}{2}) \\  \frac{2}{2}  \\ 1
(-1,1)

0:
2(2)^{0} \\ 2(1) \\ 2
(0, 2)

1:
2(2)^{1}  \\ 2(2) \\ 4
(1, 4)

2:
2(2)^{2} \\ 2(4) \\ 8
(2, 8)

3:
2(2) ^{3}  \\ 2(8) \\ 16
(3,16)

b.
To graph the equation, simply go through the points (-2, 0.5), (-1, 1), (0,2), (1,4), (2,8), and (3,16). Make sure you never go below 0 on the x-axis, because there's an asymptote there.

Hope this helps!
8 0
2 years ago
Read 2 more answers
Optimization problem: what are the dimensions of the lightest open-top right circularcylindrical can that will hold a volume of
vivado [14]
So Volume of cylinder is pi*r^2*h = 1,000 

Then lightest one means you have the smallest surface area. Which is one base and then the area of the surface. SA = pi*r^2 + 2pi*r*h 

So now you have 2 equations, so: 

h = 1,000/(pi*r^2) 
So then SA = pi*r^2 + 2pi*r*(1,000/(pi*r^2) = pi*r^2 + 2,000/r 

Derivative of SA is then 2pi*r -2,000/r^2. Set to 0 

2pi*r-2,000/r^2 =0 --> 2pi*r^3 = 2,000 --> r^3 = 1,000/pi --> r = 10/pi^(1/3) 

Now go back to the volume function: pi*r^2*h =1,000 --> 1,000/(pi*100/pi^(2/3)) = h 
<span>h = 10 / pi^(1/3)</span>
3 0
3 years ago
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