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3 years ago

There are 15 men and 20 women participating

Mathematics

2 answers:

Helen [10]3 years ago

6 0

It’s a 3 to 4 ratio of men to women

o-na [289]3 years ago

3 0

Answer:

3 men to 4 woman

Step-by-step explanation:

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Simplify. −|11|=___ -11 -121 121 11

OleMash [197]

Assignment: $\bold{Solve \ Equation: \ -\left|11\right|}$

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Answer: $\bold{-11}$

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Explanation: $\downarrow\downarrow\downarrow$

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[ Step One ] Apply Absolute Rule
Note: $\bold{Absolute \ Rule: \left|a\right|=a,\:a\ge 0}$

$\bold{\left|11\right|=11}$

[ Step Two ] Simplify

$\bold{-11}$

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$\bold{\rightarrow Mordancy \leftarrow}$

6 0

3 years ago

Read 2 more answers

If m Answer choices are - Which property is shown Reflexive Substitution Symmetric Transitive

ss7ja [257]

Answer: symmetric or reflexive

Step-by-step explanation:

I'm not forsure but I believe that these are the two options bc the other two aren't forsure sorry dude I'm trying :0

7 0

3 years ago

A four-digit number N leaves a remainder of 10 when divided by 21, a remainder of 11 when divided by 23, and a remainder of 12 w

aksik [14]

Answer:

Step-by-step explanation:

There may be a formal way to solve simultaneous Diophantine equations, but I don't know what it is. So, I addressed this by solving them 2 at a time.

First of all, we want to find a relation between integers p and q such that ...

21p+10 = 23q +11

21p -23q = 1 . . . . . . subtract 23q+10

Using the extended Euclidean algorithm or trial and error or astute observation, we find that (p, q) = (11, 10) is a solution to this equation. Then we can write p and q as ...

p = 23n+11
q = 21n+10

for some integer n.

Then our original numbers become ...

21p+10 = 21(23n+11)+10 = 483n +241 . . . . for some integer n

For this to be a 4-digit number, we must have n such that

1000 ≤ 483n +241 ≤ 9999

1.6 < n < 20.2

Modulo 25, the number (483n+241) is ...

8n +16

and we want that value to be 12:

(8n +16) mod 25 = 12

(8n +4) mod 25 = 0 . . . . . subtract 12

4(2n+1) mod 25 = 0 . . . . factor out 4

For this to be true, we must have 2n+1 be a multiple of 25. The only value of n that is in the required interval [2, 20] is n=12.

When n=12, 483n+241 = 6037. The sum of digits of 6037 is 16.

3 0

3 years ago

Find the answer for x

Ber [7]

Answer:

I think the answer is 75.

Step-by-step explanation:

because 35 + 70 is equal to 105. So now we will just subtract 105 by 180. which will give us 75.

5 0

3 years ago

Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.

const2013 [10]

Answer:

$T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

Step-by-step explanation:

Remember that for any curve $r(t)$

The tangent vector is given by

$T(t) = \frac{r'(t) }{| r'(t)| }$

And the normal vector is given by

$N(t) = \frac{T'(t)}{|T'(t)|}$

For this case, using the chain rule

$r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\$

And also remember that

$|r'(t)| = \sqrt{(-10*2tsin(t^2))^2 + ( 10*2t cos(t^2) )^2} \\\\ = \sqrt{400 t^2*( sin(t^2)^2 + cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t$

Therefore

$T(t) = r'(t) / |r'(t) | = ( -10*2tsin(t^2) , 10*2t cos(t^2) )/ 20t\\\\ = ( -10*2tsin(t^2)/ 20t , 10*2t cos(t^2) / 20t )\\= ( -sin(t^2), cos(t^2) )$

Similarly, using the quotient rule and the chain rule

$T'(t) = ( -2t cos(t^2) , -2t sin(t^2))$

And also

$|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t$

Therefore

$N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

Notice that

1. $|N(t)| = |T(t) | = \sqrt{ cos(t^2)^2 + sin(t^2)^2 } = \sqrt{1} = 1$

2. $N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0$

Simlarly

$r'(t) = (2t,-6,0) \\$

and

$|r'(t)| = \sqrt{(2t)^2 + 6^2} = \sqrt{4t^2 + 36}$

Therefore

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

Then

$T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)$

and also

$|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 + ( (3 t)/(9 + t^2)^{3/2})^2 + 0^2 }\\= 3/(t^2 + 9 )$

And since

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

6 0

3 years ago