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makkiz [27]
3 years ago
8

Which situation results in a final value of zero?

Mathematics
1 answer:
kipiarov [429]3 years ago
4 0

Answer:

B

Step-by-step explanation:

The profit made is equal to 2.25-2.25 which is 0.

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How can a number line be used to explain why 11.6 > 11.3
Angelina_Jolie [31]
In a number line its shows u were the spt of the numbers are soo. Lets say the number line goes by 2s. Its starts in 8 soo its 8, 10, 12. Go closer and its 10 , 11 , 12 . closer. 11, 11.1 , 11. 2. Untilit keeps going and u can see that 11.6 is farther to the right so that means that it is greater
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3 years ago
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A store sells almonds for $7 per pound, cashews for $10 per pound, and walnuts for $12 per pound. A customer buys 12 pounds of m
shepuryov [24]

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b

Step-by-step explanation:

I did it on edge 2021

5 0
2 years ago
Which expressions are equivalent to 2^5x2^4? Check all that apply
KIM [24]

Answer:

Step-by-step explanation:

2^9

2^-2 * 2^11

(2*2*2*2*2)*(2*2*2*2)

8 0
3 years ago
90 + 12 = 6(15 + 12) Yes or No
quester [9]

Answer:

This is false.

Step-by-step explanation:

90 +  12 = 102.

6(15+12) equals to 6(27) which is 162.


8 0
3 years ago
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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}

7 0
2 years ago
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