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OlgaM077 [116]
4 years ago
13

Which equation can be used to find the solution of (12)2x=32 ?

Mathematics
2 answers:
SOVA2 [1]4 years ago
5 0
(12)2x=32
2x= 32/12
2x=8/3

I apologize I'm not sure where they get the 5. So I'm very interested in this answer as well.


Lena [83]4 years ago
3 0

Answer: The correct option is D, i.e., -2x=5.

Explanation:

The given equation is,

(\frac{1}{2} )^{2x}=32

(\frac{1}{2} )^{2x}=2^5

Using the exponent rule (\frac{1}{a}) ^b=a^{-b}, the above equation can be written as,

2^{-2x}=2^5

On comparing both sides we get the base is same so power must be equal.

-2x=5

Therefore the required equation is -2x=5 and option D is correct.

You might be interested in
28/7 in decimal form?
lianna [129]
The answer is 4, which is an integer
4 0
3 years ago
Read 2 more answers
What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
Natasha2012 [34]

Answer:

x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

2x^2-10x-3

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

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

General form of quadratic equation is ax^2+bx+c

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

D=\sqrt{(-10)^2-4(2)(-3)}

\Rightarrow D=\sqrt{100+24}=\sqrt{124}

Now substituting D in  x=\frac{b^2\pm\sqrt{D}}{2a} we get

x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}

x=\frac{100\pm\sqrt{124}}{4}

x=\frac{100\pm2\sqrt{31}}{4}

x=\frac{50\pm\sqrt{31}}{2}

Therefore, x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}



5 0
3 years ago
Read 2 more answers
Part 1
bearhunter [10]

Answer:

Sale price:$29.02

Step-by-step explanation:

32.5/100=0.325x43.00=13.975/13.98(rounded)

43.00-13.98=29.02

4 0
3 years ago
Find the slope of (0, 4), (− 5, 2)
vredina [299]

Answer:

⅖ is the slope

Step-by-step explanation:

To find slope:

\frac{ {y}^{2} -  {y}^{1}  }{ {x}^{2} -  {x}^{1}  }

\frac{2 -4}{ - 5 - 0}

\frac{ - 2}{ - 5}

\frac{2}{5}

7 0
3 years ago
Can you help me solve for y 7/2y-8=-1
Andre45 [30]
The answer is 2
You multiply both sides by 2 (7y-16=-2)
Then move the constant to the right (7y=-2+16)
Then calculate the sum (7y=14)
The divide both sides by 7
Y=2
6 0
3 years ago
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