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mixer [17]
3 years ago
6

13+7 POINTS!!! PLZ HELP WILL GIVE BRAINLIEST!!!

Mathematics
1 answer:
adelina 88 [10]3 years ago
6 0
If x=-5 (-5)2+2=27
likewise put -4 -3 -2 -1 0 1 2 3
and in another equation also.
You might be interested in
Find x, y, and z such that x³+y³+z³=k, for each k from 1 to 100.​
love history [14]

Answer:

x3+y3+z3=k  with k is integer from 1 to 100

solution x=0 , y=0 and z=1 and k= 1

For K= 1 , we have the following solutions (x,y,x) = (1,0,0) ; or (0,1,0) ; or (0,0,1) ,

For k =1 also (9,-8,-6) or (9,-6,-8) or (-8,-6,9) or (-8,9,-6) or (-6,-8,9) or (-6,9,8)

And (-1,1,1) or (1,-1,1)

=>(x+y)3−3x2−3xy2+z3=k

=>(x+y+z)3−3(x+y)2.z−3(x+y).z2=k

=>(x+y+z)3−3(x+y)z[(x+y)−3z]=k

lety=αand z=β

=>x3=−α3−β3+k

For k= 2 we have (x,y,z) = (1,1,0) or (1,0,1) or (0,1,1)

Also for (x,y,z) = (7,-6,-5) or (7,-6,-5) or (-6,-5,7) or (-6,7,-5) or (-5,-6,7) or (-5,7,-6)

For k= 3 we have 1 solution : (x,y,z) = (1,1,1)

For k= 10 , we have the solutions (x,y,z) = (1,1,2) or (1,2,1) or (2,1,1)

For k= 9 we have the solutions (x,y,z) = (1,0,2) or (1,2,0) or (0,1,2) or (0,2,1) or (2,0,1) or (2,1,0)

For k= 8 we have (x,y,z) = ( 0,0,2) or (2,0,0) or (0,2,0)

For k= 17 => (x,y,z) = (1,2,2) or (2,1,2) or ( 2,2,1)

For k = 24 we have (x,y,z) = (2,2,2)

For k= 27 => (x,y,z) = (0,0,3) or (3,0,0) or (0,3,0)

for k= 28 => (x,y,z) = (1,0,3) or (1,3,0) or (1,3,0) or (1,0,3) or (3,0,1) or (3,1,0)

For k=29 => (x,y,z) = (1,1,3) or (1,3,1) or (3,1,1)

For k = 35 we have (x,y,z) = (0,2,3) or (0,3,2) or (3,0,2) or (3,2,0) or 2,0,3) or (2,3,0)

For k =36

we have also solution : x=1,y=2andz=3=>

13+23+33=1+8+27=36 with k= 36 , we have the following

we Have : (x, y,z) = (1, 2, 3) ; (3,2,1); (1,3,2) ; (2,1,3) ; (2,3,1), and (3,1,2)

For k= 43 we have (x,y,z) = (2,2,3) or (2,3,2) or (3,2,2)

For k = 44 we have ( 8,-7,-5) or (8,-5,-7) or (-5,-7,8) or ( -5,8,-7) or (-7,-5,8) or (-7,8,-5)

For k =54 => (x,y,z) = (13,-11,-7) ,

for k = 55 => (x,y,z) = (1,3,3) or (3,1,3) or (3,1,1)

and (x,y,z) = (10,-9,-6) or (10,-6,-9) or ( -6,10,-9) or (-6,-9,10) or (-9,10,-6) or (-9,-6,10)

For k = 62 => (x,y,z) = (3,3,2) or (2,3,3) or (3,2,3)

For k =64 => (x,y,z) = (0,0,4) or (0,4,0) or (4,0,0)

For k= 65 => (x,y,z) = (1,0,4) or (1,4,0) or (0,1,4) or (0,4,1) or (4,1,0) or (4,0,1)

For k= 66 => (x,y,z) = (1,1,4) or (1,4,1) or (4,1,1)

For k = 73 => (x,y,z) = (1,2,4) or (1,4,2) or (2,1,4) or (2,4,1) or (4,1,2) or (4,2,1)

For k= 80=> (x,y,z)= (2,2,4) or (2,4,2) or (4,2,2)

For k = 81 => (x,y,z) = (3,3,3)

For k = 90 => (x,y,z) = (11,-9,-6) or (11,-6,-9) or (-9,11,-6) or (-9,-6,11) or (-6,-9,11) or (-6,11,-9)

k = 99 => (x,y,z) = (4,3,2) or (4,2,3) or (2,3,4) or (2,4,3) or ( 3,2,4 ) or (3,4,2)

(x,y,z) = (5,-3,1) or (5,1,-3) or (-3,5,1) or (-3,1,5) or (1,-3,5) or (1,5,-3)

=> 5^3 + (-3)^3 +1 = 125 -27 +1 = 99 => for k = 99

For K = 92

6^3 + (-5)^3 +1 = 216 -125 +1 = 92

8^3 +(-7)^3

Step-by-step explanation:

4 0
3 years ago
3<br> Describe and correct the error a student made when graphing the linear equation y = -3/4x -6
likoan [24]

Answer:

In step 1 the y-intercept should be plotted at (0,-6)

Step-by-step explanation:

Remember that to find the Y intercept in any linear equation you need to use 0 as your X value, this means taking the formula in the y=mx+b form and replacing X with a 0.

Since the formula is y = -3/4x -6

We just insert a 0 insted of the "x"

y = -3/4(0) -6

y=0-6

y=6

So the y-intercept sould be placed in (0,-6)

That's what he did wrong when graphing the equation.

5 0
4 years ago
What is the code? Thank you! Whoever answers this, I will give brainliest!
shepuryov [24]

Answer:

5821

Step-by-step explanation:

For the first symbol, the number is 5.

In the first problem, the digit corresponding to that symbol is 5.

In the second one, it should be 5. The area of a triangle is base * height / 2. The base is 15 and the height is 9. Therefore, the area will be 15 * 9 / 2 = 67.5 square units. Now, the number corresponding to that symbol is 5.

In the fourth problem, the number corresponding to the symbol is also 5.

For the second symbol, the number is 8.

In the 5th problem, the number corresponding to that symbol is 8.

In the 7th problem, the number corresponding to the symbol is 8.

In the last problem, 96 is the correct answer, and everything is in the right place. I assume this must be the mistake of the maker of the problem.

For the third symbol, the number is 2

In the 1st and 3rd problems, the number corresponding to that symbol is 2

Finally, for the last symbol, the number is 1

In the 4th problem, there is a trapezoid. The formula for the area of a trapezoid is base 1 * base 2 / 2 * height. base 1 = 29, base 2 = 13. 13 + 29 = 42. 42 / 2 = 21. 21 * 15 = 315. Now, the 2nd digit, the one corresponding to the symbol, is 1.

In the last problem, you need to find the surface area of a figure. I would do this by splitting it up. The area for a triangle is base * height / 2. That means, the area of the triangle would be 8 * 7 / 2 = 28. There are four identical triangles, so you multiply this by 4 and get 112. Next, you find the area of the square which is 7 * 7. 7 * 7 = 49. Now, you add them together. 112 + 49 = 161. Now, the first and last digits corresponding to the symbol are both 1.

The answer is 5821.

4 0
3 years ago
What is the factorisation of x2 + 2x
nadezda [96]

The GCF: x

answer: x(x+2)

5 0
1 year ago
The mass of a textbook is about 1.25 kilograms. about how many pounds is this?
krok68 [10]
1 kilogram = 2.205 pounds
1.25 kilograms x 2.2 pounds = 2.75 pounds approximately
4 0
4 years ago
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