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emmasim [6.3K]
3 years ago
8

The volume of a box is 80 cubic feet with length x-3 and width x-1 and height x+5. What are the possible values of x? What are t

he possible dimensions?

Mathematics
1 answer:
Kryger [21]3 years ago
4 0

Answer:

  • the only possible value of x is 5
  • the dimensions are 2 × 4 × 10

Step-by-step explanation:

The cubic equation ...

  (x -3)(x -1)(x +5) = 80

has one real root: x = 5. Using that value for x, the dimensions become ...

  length = 5 - 3 = 2

  width = 5 - 1 = 4

  height = 5 + 5 = 10

The dimensions are (length, width, height) = (2, 4, 10).

_____

We cannot tell the thrust of the problem, since it has only one solution. Perhaps you're supposed to write the cubic in standard form and use the <em>Rational Root theorem</em> to find <em>possible values of x</em>. That form can be found to be ...

  (x -3)(x -1)(x +5) -80 = 0

  x³ +x² -17x -65 = 0

Descartes' rule of signs tells you there is one positive real root. The rational root theorem tells you possible rational roots are factors of 65:

  1, 5, 13, 65

We know that x must be greater than 3 (so all dimensions are positive). Thus <em>possible values of x are 5, 13, 65</em>, and we're pretty sure that 65 is way too large.

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Answer in standard form PLEASE.
matrenka [14]

0.9 x 10^4

Let’s break this down into steps.

So to start off with, you need to do 4.5/5 which = 0.9.

Now we can deal with the indices. 10^-3 / 10^-7 means we have to subtract them. Therefore, -3 - -7 = 4. Altogether, we have 0.9 x 10^4

The question states we should leave our answer in standard form.

So our answer is 0.9 x 10^4.
8 0
3 years ago
What is the solution to the system of equations?<br><br> y = 1.5x – 4<br><br> y = –x
sasho [114]
I attached a photo of the work. I hope it helps!

5 0
3 years ago
Read 2 more answers
Find the fifth roots of 243(cos 260° + i sin 260°).
Dmitry [639]

Answer:

z1=3 cos (52 + i sin 52)

z2 =3 cos (124 + i sin 124)

z3 = 3 cos (196 + i sin 196)

z4 =3 cos (268 + i sin 268)

z5= 3 cos (340 + i sin 340)

Step-by-step explanation:

To find the fifth roots of 243 (cos 260° + i sin 260°).

z ^ 1/5 = r^1/5 ( cis ( theta + 360 *k)/5)  where k=0,1,2,3,4


So the first root of 243 (cos 260° + i sin 260°)  

is z1 =  243^1/5 ( cis ( 260 + 360 *0)/5)  

          3 cis ( 260/5)

        = 3 cis (52)

        = 3 cos (52 + i sin 52)


The second root of  243 (cos 260° + i sin 260°)  

is z2 =  243^1/5 ( cis ( 260 + 360 *1)/5)  

          3 cis ( 620/5)

        = 3 cis (124)

        = 3 cos (124 + i sin 124)


The third root of  243 (cos 260° + i sin 260°)  

is z3 =  243^1/5 ( cis ( 260 + 360 *2)/5)  

          3 cis ( 980/5)

        = 3 cis (196)

        = 3 cos (196 + i sin 196)


The fourth root of  243 (cos 260° + i sin 260°)  

is z4 =  243^1/5 ( cis ( 260 + 360 *3)/5)  

          3 cis ( 1340/5)

        = 3 cis (268)

        = 3 cos (268 + i sin 268)


The fifth root of  243 (cos 260° + i sin 260°)  

is z5 =  243^1/5 ( cis ( 260 + 360 *4)/5)  

          3 cis ( 1700/5)

        = 3 cis (340)

        = 3 cos (340 + i sin 340)

6 0
3 years ago
Plz need help no guessing
tresset_1 [31]

C. y=-2x+1

You can test this by inputting each value of X into that equation, noticing that the Y value is always correct when you use that equation.

3 0
3 years ago
Correct answer gets Brainliest + 15 points! , help me if u can pls.
xxTIMURxx [149]
Point form: (4,5)
Equation form: x=4, y=5
3 0
3 years ago
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