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Sergeu [11.5K]
3 years ago
11

What is 11.45 as a fraction

Mathematics
2 answers:
aleksandrvk [35]3 years ago
8 0
It would be 11 9/20  or  299/20 as an improper (top-heavy) fraction. Hope this helps! XD
Viefleur [7K]3 years ago
6 0
11.45

the 11 is ur whole number...it stays ur whole number
the 0.45 is ur fraction...and since the last number (5) is in the hundredths place, put it over 100
so u now have : 11 45/100 which reduces to 11 9/20 and if u need it in improper fraction form, it is 229/20
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You run one lap around a mile track every 8 minutes. Your friend runs around the same track every 10 minutes. You both start at
Ber [7]

Answer:

Step-by-step explanation:

first runs n+1 laps

and second runs n laps in the same time.

so (n+1)8=10n

10n=8n+8

10n-8n=8

2n=8

n=8/2=4

one lap=1 mile

4 laps=4 miles

4+1=5 laps=5 miles.

first runs 5 miles whereas second runs=4 miles.

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3 years ago
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2x(-5x-3)

2x(-5x)= -10x    and 2x(-3)= -6x

= -10x sq - 6x

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3 years ago
Find k so that the following function is continuous:<br> f(x)={kx8x2if0≤x&lt;5if5≤x.
tankabanditka [31]

Check the one-sided limits:

\displaystyle \lim_{x\to5^-}f(x) = \lim_{x\to5}kx = 5k

\displaystyle \lim_{x\to5^+}f(x) = \lim_{x\to5}8x^2 = 200

If <em>f(x)</em> is to be continuous at <em>x</em> = 5, then these two limits should have the same value, which means

5<em>k</em> = 200

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Y<br> 45°<br> 30°<br> N<br> 17<br> I need help pls
melisa1 [442]

Answer:

x=29.4

Step-by-step explanation:

To help find the sides of a triangle, I use SOHCAHTOA.

SOH - CAH - TOA

S=O/H (Sine=\frac{opposite}{hypotenuse} )

C=A/H (Cosine=\frac{adjacent}{hypotenuse} )

T=O/A (Tangent =\frac{opposite}{adjacent} )

Let's find x first. We know 2 parts of the triangle: the adjacent leg, and the angle that is opposite of that.

So it would be tangent (tan).

opposite = 17

adjacent = 30°

Plug these numbers into the calculator and you should get your answer.

\frac{17}{tan(30)} = 29.44486372...

x=29.4

Try to do <em>y</em> on your own and let me know if you have any questions!

Also correct me if I'm wrong on anything.

Hope this information helps!

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