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Amiraneli [1.4K]
2 years ago
6

Many of you will live to the year 2100, which is not a leap year. Not all years which are divisible by 4 are leap years. A cente

nnial year (multiple of 100) is a leap year only if it is not divisible by 400. So 2000 is (was) a leap year, 2400 is a leap year, but 2100 and 2200 are not leap years. Non-centennial years which are multiples of 4 are always leap years. If P is the statement
and Q is the statement and R is the statement
P: ”L mod 4 = 0”
Q: ”L mod 100 = 0” R: ”L mod 400 = 0”
Mathematics
1 answer:
Slav-nsk [51]2 years ago
3 0

Answer:

For example, the years 1600, 2000, and 2400 are century leap years since those numbers are divisible by 400, while 1700, 1800, 1900, 2100, 2200, and 2300 are common years despite being divisible by 4.

Step-by-step explanation:

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How many miles did Brian run?
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Lucia left a tip equal to 18% of the cost of her lunch. If she left $1.53 as her tip, what was the cost of her lunch, in dollars
xxMikexx [17]

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$8.50

Step-by-step explanation:

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7 0
2 years ago
Find the vector projection of B onto A if A = 5i + 11j – 2k,B = 4i + 7k​
valkas [14]

Answer:

\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\

Step-by-step explanation:

Given A = 5i + 11j – 2k and B = 4i + 7k​, the vector projection of B unto a is expressed as proj_ab = \dfrac{b.a}{||a||^2} * a

b.a = (5i + 11j – 2k)*( 4i + 0j + 7k)

note that i.i = j.j = k.k  =1

b.a = 5(4)+11(0)-2(7)

b.a = 20-14

b.a = 6

||a|| = √5²+11²+(-2)²

||a|| = √25+121+4

||a|| = √130

square both sides

||a||² = (√130)

||a||²  = 130

proj_ab = \dfrac{6}{130} * (5i+11j-2k)\\\\proj_ab = \frac{30}{130} i+\frac{11}{130} j-\frac{12}{130} k\\\\proj_ab = \frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\\\

<em>Hence the projection of b unto a is expressed as </em>\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\<em></em>

7 0
3 years ago
Suppose that the functions r and a are defined for all real numbers x as follows. r(x)=2x-1 S(x)=5x write the expressions for (r
NeTakaya

\boxed{(r-s)(x)=-3x-1} \\ \\ \boxed{(r\cdot s)(x)=10x^2-5x} \\ \\ \boxed{(r+s)(-2)=-15}

<h2>Explanation:</h2>

In this exercise, we have the following functions:

r(x)=2x-1 \\ \\ s(x)=5x

And they are defined for all real numbers x. So we have to write the following expressions:

First expression:

(r-s)(x)

That is, we subtract s(x) from r(x):

(r-s)(x)=2x-1-5x \\ \\ Combine \ like \ terms: \\ \\ (r-s)(x)=(2x-5x)-1 \\ \\ \boxed{(r-s)(x)=-3x-1}

Second expression:

(r\cdot s)(x)

That is, we get the product of s(x) and r(x):

(r\cdot s)(x)=(2x-1)(5x) \\ \\ By \ distributive \ property: \\ \\ (r\cdot s)(x)=(2x)(5x)-(1)(5x) \\ \\ \boxed{(r\cdot s)(x)=10x^2-5x}

Third expression:

Here we need to evaluate:

(r+s)(-2)

First of all, we find the sum of functions r(x) and s(x):

(r+s)(x)=2x-1+5x \\ \\ Combine \ like \ terms: \\ \\ (r+s)(x)=(2x+5x)-1 \\ \\ (r+s)(x)=7x-1

Finally, substituting x = -2:

(r+s)(-2)=7(-2)-1 \\ \\ (r+s)(-2)=-14-1 \\ \\ \boxed{(r+s)(-2)=-15}

<h2>Learn more: </h2>

Parabola: brainly.com/question/12178203

#LearnWithBrainly

5 0
3 years ago
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