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3 years ago

How many significant digits are there in 17.0

1 answer:

4 0

The answer is three significant figures. The 1 and the 7 are both significant, because they are non-zero quantities.
This is where significant figures gets a little more complicated, because if a zero is used as a placeholder (i.e. 0.00027 cm) then it is insignificant.
But in the case above, the zero isn't being used as a placeholder, and thus, is significant.

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bixtya [17]

Answer:

12 inches

Step-by-step explanation:

Set up an equation:

Variable x = height of the scale model

3/18 = x/72

Cross multiply:

3 × 72 = 18 × x

216 = 18x

Divide both sides by 18:

4 0

3 years ago

Can someone help<br> With this math question please

zhannawk [14.2K]

The equation of a line is y = mx+b, where m is the slope. In this equation m = 3, so the slope is upward and to the right, above the 45-degree angle. Only graph J fits this description.

6 0

3 years ago

If ΔEFG ~ ΔLMN with a ratio of 2:1, which of the following is true?

strojnjashka [21]

Answer:

segment EF over segment LM equals segment FG over segment MN

Step-by-step explanation:

The triangles are similar, not congruent, so any answer choice with the word "congruent" can be ignored.

The sequence of letters in the triangle name tells you the corresponding segments:

EF corresponds to LM
EG corresponds to LN
FG corresponds to MN

Corresponding segments have the same ratio, so ...

EF/LM = FG/MN . . . . . . matches the first answer choice

EF/LM = EG/LN . . . . does not match the 3rd answer choice

7 0

3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$