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MakcuM [25]
3 years ago
6

Find two positive numbers whose difference is 2828 and whose product is 39003900.

Mathematics
1 answer:
Anastasy [175]3 years ago
7 0
If the question is:
"<span>Find two positive numbers whose difference is 28 and whose product is 3900</span>"

If the larger number is x, then
x-3900/x=28
multiply by x and transpose
x^2-28x-3900=0
Factor
(x+78)(x-50)=0
Using the zero product property,
x=-78 or x=50, ie.  x=50, y=-78

NOTE: if question requires both numbers be positive, there is no solution.
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Can you help with this?<br><br> Length x width x height
Gemiola [76]

Answer:

75/256 in³    <    64/125 in³

Step-by-step explanation:

First, find the volumes of the two 3-D figures:

For Figure 1:

Using the volume formula:

V = l × w × h

   = 6/8 × 5/8 × 5/8

   = (6/8 × 5/8) × 5/8

   = 30/64 × 5/8

V = 150/512

Simplify:

V = 75/256

So Figure 1's volume is 75/256 in³.

For Figure 2:

V = l × w × h

   = 4/5 × 4/5 × 4/5

   = (4/5 × 4/5) × 4/5

   = 16/25 × 4/5

   = 64/125

So Figure 2's volume is 64/125 in³.

Now, comparing the two volumes:

You can cross multiply the fractions to figure out which fraction is greater:

Figure 1                       Figure 2

<u> 75 </u>                =                <u> 64 </u>

256                                  125

75 (125)          =             256 (64)

9375               =             16384

So Figure 2 has the greater volume.

Your final result should be:

75/256 in³    <    64/125 in³

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Answer:

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Step-by-step explanation:

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Answer:

x=33.4

Step-by-step explanation:

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Nadusha1986 [10]

Answer:

[x+6y+2z][x²+(6y)²+(2z)²-6xy-12yz-2xz]

Step-by-step explanation:

x³+216y³+8z³-36xyz

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As we know

a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)

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[x+6y+2z][(x²+(6y)²+(2z)²-x×6y-6y×2z-x×2z]

[x+6y+2z][x²+(6y)²+(2z)²-6xy-12yz-2xz]

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