Subduction Zone
Veteran Member
196 - 182*0.5 = 7!!
Did I make any errors?
Did I make any errors?
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That is not the answer that I was looking for, but thank you.Yes. You asked me to check your math.
Yup, you made an error.196 - 182*0.5 = 7!!
Did I make any errors?
Did I make any errors?
196 - 182*0.5 = 7!!
Did I make any errors?
196 - 182*0.5 = 7!!
Did I make any errors?
Nope. I followed the order of operations.
What year in reference are we asking? Afterall, PEMDAS is today but not yesterday and probably not tomorrow.196 - 182*0.5 = 7!!
Did I make any errors?
Did you read the thread?What year in reference are we asking? Afterall, PEMDAS is today but not yesterday and probably not tomorrow.
Yes. Just throwing out the but if history of the order of operations and what we say it is today hasn't always been.Did you read the thread?
What do the exclamation points represent? Anything other than excitement in your answer?
I remember learning BODMAS at school in about 1964. My son learnt a slightly different mnemonic for the same thing in about 2014. So yeah 105.196 - 182*0.5 = 7!!
Did I make any errors?
Nah, it is the two, not just one, exclamation points.I remember learning BODMAS at school in about 1964. My son learnt a slightly different mnemonic for the same thing in about 2014. So yeah 105.
Unless you are going to tell us that the * symbol for some reason is to be applied in a different order from x, in which I look forward to your explanation.
One exclamation point: factorial. Two exclamation points: double factorial.Google Gemini choked utterly. ChatGPT tried and failed
Q: In what numeric base is this equation true 196 - 182*0.5 = 7
A: To find out in what numeric base this equation is true, let's express it generically in base bb:
196b−182b×0.5b=7b196b−182b×0.5b=7b
First, let's convert the decimal numbers to base bb:
19610=1×b2+9×b1+6×b019610=1×b2+9×b1+6×b0 18210=1×b2+8×b1+2×b018210=1×b2+8×b1+2×b0 0.510=5×b−10.510=5×b−1
Now, let's substitute these values into the equation and solve for bb:
(1×b2+9×b1+6×b0)−(1×b2+8×b1+2×b0)×(5×b−1)=7b(1×b2+9×b1+6×b0)−(1×b2+8×b1+2×b0)×(5×b−1)=7b
Simplify the equation:
(1×b2+9×b1+6×b0)−(1×b2+8×b1+2×b0)×(5×b−1)=7b(1×b2+9×b1+6×b0)−(1×b2+8×b1+2×b0)×(5×b−1)=7b (1×b2+9×b1+6×b0)−(5×b−1×(1×b2+8×b1+2×b0))=7b(1×b2+9×b1+6×b0)−(5×b−1×(1×b2+8×b1+2×b0))=7b (1×b2+9×b1+6×b0)−(5×b−1×b2+40×b0+10×b−1)=7b(1×b2+9×b1+6×b0)−(5×b−1×b2+40×b0+10×b−1)=7b 1×b2+9×b1+6×b0−5×b1−40×b0−10×b−1=7b1×b2+9×b1+6×b0−5×b1−40×b0−10×b−1=7b
Now, equate the coefficients of corresponding powers of bb on both sides of the equation:
1b2−5b1+6b0−10b−1−40b0=71b2−5b1+6b0−10b−1−40b0=7
b2−5b+6−10b−40=7b2−5b+6−b10−40=7
b2−5b−34−10b=0b2−5b−34−b10=0
This equation can be solved to find the value(s) of bb that make the original equation true. However, without further context or constraints, it's difficult to determine the exact base in which the equation holds true. It's worth noting that solutions for bb may not always be unique, and some bases may not be practical or meaningful in certain contexts.