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cestrela7 [59]
3 years ago
12

How many monday thru fridays are in a year?

Mathematics
2 answers:
Ghella [55]3 years ago
5 0
It should be about 251 days that are from monday though friday
stich3 [128]3 years ago
3 0
Without weekends, there are 260 or 261 days in a year (depends on whether it's leap year or not). There are 52.14 weeks in a year and two weekend days in each, so multiply 52.14 by 5 to get 261 (when rounded up).
Hope this helps :-)
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A sequence of transformations maps ∆ABC to ∆A′B′C′. The sequence of transformations that maps ∆ABC to ∆A′B′C′ is a reflection ac
7nadin3 [17]

Answer:

the answe is d

Step-by-step explanation:

ye

3 0
3 years ago
Given: △PRS, RS=10<br> m∠P=45°, m∠S=60°<br> Find: Perimeter of △PRS
Keith_Richards [23]

The law of sines can be used to find the missing side lengths.

... RP = sin(60°)/sin(45°)·10 = 10√(3/2) ≈ 12.247

... PS = sin(75°)/sin(45°)·10 ≈ 13.660

Then the perimeter is

... perimeter = RS + RP + PS

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3 0
4 years ago
Evaluate the limit of
tekilochka [14]

Answer:

Option a is correct -- \frac{9}{2}

Step-by-step explanation:

If we put infinity directly into the expression we get ∞ + ∞ expression. In order to circumvent it we divide both numerator and denominator with the greatest exponent.

\lim_{x \to \infty} \frac{9n^{3}+5n-2}{2n^{3}}

 Divide each term by n³

=\lim_{x \to \infty} \frac{\frac{9n^{3}}{n^{3}}+\frac{5n}{n^{3}} -\frac{2}{n^{3}}}{\frac{2n^{3}}{n^{3}} }

=\lim_{x \to \infty}\frac{9(1)+\frac{5}{n^{2}}-\frac{2}{n^{3}}}{2(1)} } \\=\lim_{x \to \infty}\frac{9+\frac{5}{n^{2}}-\frac{2}{n^{3}}}{2}} \\\\by putting n = infinity \frac{5}{n^{2}} becomes 0 and\frac{2}{n^{3}} becomes 0 \\=\frac{9+0-0}{2} \\=\frac{9}{2}



4 0
3 years ago
Diego says that the product of 0.51 times 2.427 will have five decimal places.is Diego correct?
labwork [276]
No, the decimal will be over four places
7 0
3 years ago
35% of something is 42
Ugo [173]
35%=35/100=0.35

0.35x=42
divide both sides by 0.35
x=120

35% of 120 is 42

Answer=120
4 0
3 years ago
Read 2 more answers
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