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slava [35]
3 years ago
12

Which expression is equivalent to 64y18-1000z6

Mathematics
2 answers:
loris [4]3 years ago
8 0
64y^18 -1000z^6 = (4y^6)^3 -(10z^2)^3
Tcecarenko [31]3 years ago
8 0

Answer:

64y^{18} - 1000z^{6} = (4y^{6} )^3 - (10z^2)^3 =  (4y^6 - 10z^2) (16y^{12} + 40y^6z^2 +100z^4)

Step-by-step explanation:

The given expression is 64y^{18} -1000z^{6}

Here we can rewrite the given expression.

64 can be written as 4*4*4 = 4^{3}

y^{18} =( y^{6}) ^3 using the power of power rule of exponent.

So. 64y^{18} = (4y^6)^{3}

Now we can write 1000z^{6} = (10z^2)^3

Therefore, 64y^{18} - 1000z^{6} = (4y^{6} )^3 - (10z^2)^3

This can be written as using the formula (a - b)^3 = (a - b)(a^2 + ab + b^2)

Here a = 4y^6 and b = 10z^2

= (4y^6 - 10z^2) (16y^{12} + 40y^6z^2 +100z^4)

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3 years ago
Suppose that demand in period 1 was 7 units and the demand in period 2 was 9 units. Assume that the forecast for period 1 was fo
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Answer:

Step-by-step explanation:

Forecast for period 1 is 5

Demand For Period 1 is 7

Demand for Period  2 is 9  

Forecast  can be given by

F_{t+1}=F_t+\alpha (D_t-F_t)

where

F_{t+1}=Future Forecast

F_t=Present\ Period\ Forecast

D_t=Present\ Period\ Demand

\alpha =smoothing\ constant  

F_{t+1}=5+0.2(7-5)

F_{t+1}=5.4

Forecast for Period 3

F_{t+2}=F_{t+1}+\alpha (D_{t+1}-F_{t+1})

F_{t+2}=5.4+0.2\cdot (9-5.4)

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3 years ago
Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.
Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ;  x = \frac{2-(3)\sqrt{2}}{2}

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

for the equation ax² + bx + c = 0

thus,

the roots are

x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}

or

x = \frac{6\pm\sqrt{36-12}}{2}

or

x = \frac{6+\sqrt{24}}{2} and, x = x = \frac{6-\sqrt{24}}{2}

or

x = \frac{6+2\sqrt{6}}{2} and, x = x = \frac{6-2\sqrt{6}}{2}

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}

or

x = \frac{4\pm\sqrt{16+56}}{4}

or

x = \frac{4+\sqrt{72}}{4} and, x = x = \frac{4-\sqrt{72}}{4}

or

x = \frac{4+\sqrt{2^2\times3^2\times2}}{2} and, x = x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}

or

x = \frac{4+(2\times3)\sqrt{2}}{2} and, x = x = \frac{4-(2\times3)\sqrt{2}}{4}

or

x = \frac{2+3\sqrt{2}}{2} and, x = \frac{2-(3)\sqrt{2}}{2}

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ; x = \frac{2-(3)\sqrt{2}}{2}

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