Answer:
yes
Step-by-step explanation:
As an example
(x₁, y₁ ) = (1,2) and (x₂, y₂ ) = (5,3), then
d =
=
=
= 
Now let (x₁, y₁ ) = (5,3) and (x₂, y₂ ) = (1,2), then
d =
=
=
= 
Answer:
- Perimeter = 22*sqrt(2)
- Area = 60.5 inches
- D
Step-by-step explanation:
Remark
You need 2 facts.
- A square has 4 equal sides.
- It contains (by definition) 1 right angle but since we are not including and statement about parallel sides, it needs 4 right angles.
That means you can use the Pythagorean Theorem.
If one side of a square is a then the 1 after it is a as well.
Formula
- a^2 + a^2 = c^2
- 2a^2 = c^2
Givens
Solution
- 2a^2 = 11^2
- 2a^2 = 121 Divide by 2
- a^2 = 121/2 Take the square root of both sides
- sqrt(a^2) = sqr(121/2)
- a = 11/sqrt(2) Rationalize the denominator
- a = 11 * sqrt(2)/[sqrt(2) * sqrt(2)]
- a = 11 * sqrt(2) / 2
<em><u>Perimeter</u></em>
P = 4s
- P = 4*11*sqrt(2)/2
- P = 44*sqrt(2)/2
- P = 22*sqrt(2)
You don't need the area. The answer is D
<em><u>Area</u></em>
- Area = s^2
- Area = (11*sqrt(2)/2 ) ^2
- Area = 121 * 2 / 4
- Area = 60.5
Answer:
6^-1
Step-by-step explanation:
6^-6 - (-5) = 6^(-6 + 5) = 6^(-1)
A lot of things are going on in this question. Notice how the 5 is handled. It starts off in the denominator as - 5 and when brought to the numerator it becomes -(-5) which makes it + 5.
The answer is then easy. It is 6^(-6 + 5) = 6^-1
You could give the answer as 1/6 but that is not what the question says.
Let x be the distance traveled on the highway and y the distance traveled in the city, so:
Now, the system of equations in matrix form will be:
![\left[\begin{array}{ccc}1&1&\\ \frac{1}{65} & \frac{1}{25} &\end{array}\right] \left[\begin{array}{ccc}x&\\y&\end{array}\right] = \left[\begin{array}{ccc}375&\\7&\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%26%5C%5C%20%5Cfrac%7B1%7D%7B65%7D%20%26%20%5Cfrac%7B1%7D%7B25%7D%20%26%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26%5C%5Cy%26%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D375%26%5C%5C7%26%5Cend%7Barray%7D%5Cright%5D%20)
Next, we are going to find the determinant:
![D= \left[\begin{array}{ccc}1&1\\ \frac{1}{65} & \frac{1}{25} \end{array}\right] =(1)( \frac{1}{25}) - (1)( \frac{1}{65} )= \frac{8}{325}](https://tex.z-dn.net/?f=D%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C%20%5Cfrac%7B1%7D%7B65%7D%20%26%20%5Cfrac%7B1%7D%7B25%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%281%29%28%20%5Cfrac%7B1%7D%7B25%7D%29%20-%20%281%29%28%20%5Cfrac%7B1%7D%7B65%7D%20%29%3D%20%5Cfrac%7B8%7D%7B325%7D%20)
Next, we are going to find the determinant of x:
![D_{x} = \left[\begin{array}{ccc}375&1\\7& \frac{1}{25} \end{array}\right] = (375)( \frac{1}{25} )-(1)(7)=8](https://tex.z-dn.net/?f=%20D_%7Bx%7D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D375%261%5C%5C7%26%20%5Cfrac%7B1%7D%7B25%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%28375%29%28%20%5Cfrac%7B1%7D%7B25%7D%20%29-%281%29%287%29%3D8)
Now, we can find x:

Now that we know the value of x, we can find y:

Remember that time equals distance over velocity; therefore, the time on the highway will be:

An the time on the city will be:

We can conclude that the bus was five hours on the highway and two hours in the city.
it's A
Step-by-step explanation:
describes 55 per unit/the hour