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2 years ago

Why must a given measurement always be reported to the correct number of significant figures.

1 answer:

5 0

The simple reason why it is very important for a given measurement always be reported to the correct number of significant figures is because solved answers usually, frequently and most of the time depend on the number of significant figures in the values used during the calculation

<h3>Significant figures</h3>

There are accurate figures which is used in solving mathematical problems

So therefore, the simple reason why it is very important for a given measurement always be reported to the correct number of significant figures is because solved answers usually, frequently and most of the time depend on the number of significant figures in the values used during the calculation

Learn more about significant figures:

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Answer please thanks for answer

zavuch27 [327]

Answer: 15.8

In this number, .8 is the tenths place. So round it to this number.

.839 rounded to nearest tenth = .8

Your result should be 15.8.

15.839 rounded to the nearest tenth is 15.8.

8 0

3 years ago

Is 4 √ 7 2 written in simplest radical form<br><br> yes or no

lord [1]

Answer:

No.

Step-by-step explanation:

Notice that 72 can be written as 8 * 9.

4√72 = 4(√8)(√9) = 4(√8)(3) = 12√8.

Hence the answer is No.

4 0

3 years ago

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If an artist has 14 pounds of clay and she uses 5 1/4 pounds. How many pounds of clay will she have left?

Licemer1 [7]

Answer: She will have 1.75 pounds of clay left

Step-by-step explanation: you need to divide 51/4 and it is 12.75 and then you subtract 14-12.75 and get = 1.25

6 0

3 years ago

Mia works at a smoothie shop. For h hours of work, Mia earns 8h dollars. Last weekend, Mia worked 9 hours.

anastassius [24]

Answer:

72$

Step-by-step explanation:

h=9

8*9=72$

8 0

3 years ago

Can someone please explain this to me? Thanks!

Makovka662 [10]

Answer: Choice D

$\displaystyle F\ '(x) = 2x\sqrt{1+x^6}\\\\$

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Explanation:

Let g(t) be the antiderivative of $g'(t) = \sqrt{1+t^3}$ . We don't need to find out what g(t) is exactly.

Recall by the fundamental theorem of calculus, we can say the following:

$\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)$

This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).

So,

$\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)\\\\$

From here, we apply the derivative with respect to x to both sides. Note that the $g(\pi)$ portion is a constant, so $g'(\pi) = 0$

$\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}\\\\$

$\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}\\\\$

Answer is choice D

5 0

2 years ago

Read 2 more answers