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Katyanochek1 [597]
2 years ago
15

What is 6 in 16.5 rounded to

Mathematics
1 answer:
Delicious77 [7]2 years ago
7 0

Answer:

The 6 in 16.5 will be rounded to 7.  

So 16.5 would become 17.

Step-by-step explanation:

You might be interested in
Problem page rita drive 420 miles using 18 gallons of gas. at this rate, how many gallons of gas would she need to drive 357 mil
bulgar [2K]
The answer is 8,330 
because you have to take 420 and divied it by 18 and youll get 23.33333 then you times it to 357 

hope you are satisfied 
3 0
3 years ago
Calculate the limit values:
Nataliya [291]
A) This particular limit is of the indeterminate form,
\frac{ \infty }{ \infty }
if we plug in infinity directly, though it is not a number just to check.

If a limit is in this form, we apply L'Hopital's Rule.

's
Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_ {x \rightarrow \infty } \frac{( ln(x ^{2} + 1 ) ) '}{x ' }
So we take the derivatives and obtain,

Lim_ {x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ \frac{2x}{x^{2} + 1} }{1}

Still it is of the same indeterminate form, so we apply the rule again,

Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 2 }{2x}

This simplifies to,

Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 1 }{x} = 0

b) This limit is also of the indeterminate form,

\frac{0}{0}
we still apply the L'Hopital's Rule,

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ (tanx)'}{x ' }

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (x) }{1 }

When we plug in zero now we obtain,

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (0) }{1 } = \frac{1}{1} = 1
c) This also in the same indeterminate form

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ ({e}^{2x} - 1 - 2x)'}{( {x}^{2} ) ' }

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (2{e}^{2x} - 2)}{ 2x }

It is still of that indeterminate form so we apply the rule again, to obtain;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (4{e}^{2x} )}{ 2 }

Now we have remove the discontinuity, we can evaluate the limit now, plugging in zero to obtain;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = \frac{ (4{e}^{2(0)} )}{ 2 }

This gives us;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } =\frac{ (4(1) )}{ 2 }=2

d) Lim_ {x \rightarrow +\infty }\sqrt{x^2+2x}-x

For this kind of question we need to rationalize the radical function, to obtain;

Lim_ {x \rightarrow +\infty }\frac{2x}{\sqrt{x^2+2x}+x}

We now divide both the numerator and denominator by x, to obtain,

Lim_ {x \rightarrow +\infty }\frac{2}{\sqrt{1+\frac{2}{x}}+1}

This simplifies to,

=\frac{2}{\sqrt{1+0}+1}=1
5 0
3 years ago
Mr Prothero places £3003 into a
mamaluj [8]

Answer:

£3,543.54

Step-by-step explanation:

Amount after 2 years = Principal + interest

Get the interest

Interest = PRT/100

Interest = 3003 * 9 * 2/100

Interest = 3003 * 18/100

Interest  = 30.03 * 18

Interest  = 540.54

Amount after 2 years = 3003 + 540.54

Amount after 2 years = £3,543.54

6 0
3 years ago
What is the missing table value?
Digiron [165]

Answer:

10.16

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
GI bisects DGH so that m DGI is x-3 and mIGH is 2x-13 find the value of x
baherus [9]

x = 10

since GI bisects ∠DGH then ∠DGI = ∠IGH, hence

2x - 13 = x - 3 ( subtract x from both sides )

x - 13 = - 3 ( add 13 to both sides )

x = 10



8 0
2 years ago
Read 2 more answers
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