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Dafna11 [192]
3 years ago
7

How many kilometers they traveled by bike

Mathematics
1 answer:
ad-work [718]3 years ago
5 0
So if you divide the total in half its 162.5. So if they traveled 75km more by bike, you add 75 to that which equals 237.5. So your answer is 237.5
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Hya think of a number "m" . She square it, add "y" to the answer and then subtract 7 from it. The final answer is "A". 1) Derive
Bond [772]
<h2><u>Solution (1)</u> :</h2>

Given, to find A we have to :

  • square m
  • Add y to m²
  • Subtract 7 from m² + y

From the question, the following equation can be formed :

=\tt  A = {m}^{2}   + y - 7

Therefore, the formula for finding A = m² + y - 7

<h2><u>Solution (2)</u> :</h2>

The value of A we can derive from the formula is :

=\tt A =  {m}^{2}  + y - 7

Value of m = 3 (given)

Which means :

=\tt A =   {3}^{2}  + y  - 7

=\tt \: A = 9 + y - 7

=\tt A = 9 - 7 + y

\color{plum} =\tt A = 2 + y

Thus, the value of A = 2+y

Therefore, the value of A = <u>2+y</u>

7 0
2 years ago
How do you get an equation to solve for all 3 sides of a triangle?
slavikrds [6]

In your solving toolbox (along with your pen, paper and calculator) you have these 3 equations:

1. The angles always add to 180°:

A + B + C = 180°

When you know two angles you can find the third.

 

2. Law of Sines (the Sine Rule):

Law of Sines

When there is an angle opposite a side, this equation comes to the rescue.

Note: angle A is opposite side a, B is opposite b, and C is opposite c.

 

3. Law of Cosines (the Cosine Rule):

Law of Cosines

This is the hardest to use (and remember) but it is sometimes needed  

to get you out of difficult situations.

It is an enhanced version of the Pythagoras Theorem that works  

on any triangle.

With those three equations you can solve any triangle (if it can be solved at all).

Six Different Types (More Detail)

There are SIX different types of puzzles you may need to solve. Get familiar with them:

1. AAA:

This means we are given all three angles of a triangle, but no sides.

AAA Triangle

AAA triangles are impossible to solve further since there are is nothing to show us size ... we know the shape but not how big it is.

We need to know at least one side to go further. See Solving "AAA" Triangles .

 

2. AAS

This mean we are given two angles of a triangle and one side, which is not the side adjacent to the two given angles.

AAS Triangle

Such a triangle can be solved by using Angles of a Triangle to find the other angle, and The Law of Sines to find each of the other two sides. See Solving "AAS" Triangles.

 

3. ASA

This means we are given two angles of a triangle and one side, which is the side adjacent to the two given angles.

ASA Triangle

In this case we find the third angle by using Angles of a Triangle, then use The Law of Sines to find each of the other two sides. See Solving "ASA" Triangles .

 

4. SAS

This means we are given two sides and the included angle.

SAS Triangle

For this type of triangle, we must use The Law of Cosines first to calculate the third side of the triangle; then we can use The Law of Sines to find one of the other two angles, and finally use Angles of a Triangle to find the last angle. See Solving "SAS" Triangles .

 

5. SSA

This means we are given two sides and one angle that is not the included angle.

SSA Triangle

In this case, use The Law of Sines first to find either one of the other two angles, then use Angles of a Triangle to find the third angle, then The Law of Sines again to find the final side. See Solving "SSA" Triangles .

 

6. SSS

This means we are given all three sides of a triangle, but no angles.

SSS Triangle

In this case, we have no choice. We must use The Law of Cosines first to find any one of the three angles, then we can use The Law of Sines (or use The Law of Cosines again) to find a second angle, and finally Angles of a Triangle to find the third angle.

7 0
3 years ago
Read 2 more answers
For each given p, let ???? have a binomial distribution with parameters p and ????. Suppose that ???? is itself binomially distr
pshichka [43]

Answer:

See the proof below.

Step-by-step explanation:

Assuming this complete question: "For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed with parameters q and M. Formulate Z as a random sum and show that Z has a binomial distribution with parameters pq and M."

Solution to the problem

For this case we can assume that we have N independent variables X_i with the following distribution:

X_i Bin (1,p) = Be(p) bernoulli on this case with probability of success p, and all the N variables are independent distributed. We can define the random variable Z like this:

Z = \sum_{i=1}^N X_i

From the info given we know that N \sim Bin (M,q)

We need to proof that Z \sim Bin (M, pq) by the definition of binomial random variable then we need to show that:

E(Z) = Mpq

Var (Z) = Mpq(1-pq)

The deduction is based on the definition of independent random variables, we can do this:

E(Z) = E(N) E(X) = Mq (p)= Mpq

And for the variance of Z we can do this:

Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2

Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2

And if we take common factor Mpq we got:

Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq]

And as we can see then we can conclude that   Z \sim Bin (M, pq)

8 0
3 years ago
Determine whether AB ← → and CD ← → − are parallel, perpendicular, or neither. A(-6, 2), B(−3, -4), C(1, -3), D(3, 4)
guapka [62]
Neither they connect but don’t make a 90 degree angle

6 0
2 years ago
PLSSSSSS HELP
Anton [14]

Answer:

The answer is D. (3+22)÷5

5 0
3 years ago
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