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SIZIF [17.4K]
2 years ago
5

Please help answer! :) How are significant digits related to calculations using measurements?

Mathematics
1 answer:
Marysya12 [62]2 years ago
3 0

The number of significant figures is the number of digits that is believed to be correct by the person doing the measuring.

<h3>What are significant figures?</h3>

It should be noted that significant figures are the number of digits in a value, that contribute to the degree of accuracy of the value. We start counting significant figures at the first non-zero digit.

Precision refers to how closely individual measurements agree with each other. In any measurement, the number of significant figures is critical.

Learn more about significant figures on:

brainly.com/question/24491627

#SPJ1

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List the the single -digits factor of 924.
Natalka [10]
1,2,3,4,6,7,8. This is the answer
3 0
3 years ago
Nicole invested her savings in two investment funds. The amount she invested in Fund A was $3000 less than the amount she invest
victus00 [196]
<span>Let us denote the amount of investment in the Fund B with x. We know that Fund B returned 8%  and that in Fund A Nicole invested </span>$3000 less than the amount she invested in Fund B, so we can write: x-3000<span>Fund A returned 6%. 
Interest = 2060
The percent has been converted by multiplying the equation by 100 
8x+6(x-3000)= 2060*100 
8x+6x-18000=206000
</span><span>14x=206000+18000
14x=224000
x=16000
Nicole invested in Fund B $16000 and in Fund A </span><span> $16000-3000=13000.</span><span>

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8 0
3 years ago
ANSWER ASAP PLEAZE AND SHOW WORK
Gwar [14]

Answer: x^(5/6)

Step-by-step explanation: screenshot

5 0
2 years ago
If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?
Naddik [55]

Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor  (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2

where the imaginary unit has disappeared, making the expression real.

So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

5 0
3 years ago
How many interior angles does a triangle have
Aleonysh [2.5K]
There are 3 interior angles.
7 0
3 years ago
Read 2 more answers
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