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Dafna11 [192]
2 years ago
12

A person bikes 2

Mathematics
1 answer:
Lerok [7]2 years ago
5 0

Answer:

The person bikes 20 miles in an hour

Step-by-step explanation:

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sasho [114]
$12 x .30= $3.60 mark up so new price was $15.60.

$15.60 x .25= $3.90 discount, so the new price with discount would be $11.70
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insens350 [35]

Answer:

<em>Given-</em> y= -5x^2 +5

ATQ,

if it translated vertically downwards then the new function will be,

h = y - 5 = -5x^2

<u>PLEASE MARK AS BRAINLIEST AND FOLLOW IF IT HELPED YOU!</u>

4 0
3 years ago
The school cafeteria is having "Make your own Sandwich" day. You have a choice of white, rye, or oat bread; turkey, harm or chic
Oksanka [162]

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Most likely 1/3 (one third) because there are three options to choose from, Ham, turkey of chicken.

4 0
4 years ago
Randy is walking home from school. According to the diagram above, what is his total distance from school to home? Show your wor
In-s [12.5K]

Answer:

First, when he walks, we can see in the image that between the school and his house he must walk 4 times a distance of 0.5km, so this is a total of 4¨*0.5km = 2km.

Then he needs to walk 2km.

Now if he has a jet-pack, he can ignore the buildings and just take the shorter path, here we can draw a triangle rectangle, in such a way that the hypotenuse of this triangle is the distance between the home and the school.

One of the catheti is the vertical distance (two blocks of 0.5km, so this catheti has a length of 2*0.5km = 1km), and the other one is the horizontal distance, also 1km.

The actual distance of this path is given by the Pythagorean's theorem:

A^2 + B^2 = H^2

Where A and B are the cathetus, and H is the hypotenuse, then:

H^2 = 1km^2 + 1km^2

H = (√2)km = 1.41km.

Now, in the case that he has a jet-pack, he can actually go to the school using this hypotenuse line as his path, in this case the distance and the displacement would be the same.

This is because the definitions of distance and displacement are:

Distance: "how much ground an object has covered"

Displacement: "Difference between the final position and the initial position"

When he walks, the distance is 2km and the displacement is 1.41km , but when he uses the jet pack, the distance is equal to the displacement, both are 1.41km.

7 0
4 years ago
Apply Crammer's Rule to find the solution to the following quations .2x + 3y = 1, 3x + y = 5​
Bond [772]

Answer:

The solution to the equation system given is:

  • <u>x = 2</u>
  • <u>y = -1</u>

Step-by-step explanation:

First, we must know the equations given:

  1. 2x + 3y = 1
  2. 3x + y = 5​

Following Crammer's Rule, we have the matrix form:

\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] =\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}1\\5\end{array}\right]

Now we solve using the determinants:

x=\frac{\left[\begin{array}{ccc}1&3\\5&1\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(1*1)-(5*3)}{(2*1)-(3*3)} = \frac{1-15}{2-9} =\frac{-14}{-7} = 2

y=\frac{\left[\begin{array}{ccc}2&1\\3&5\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(2*5)-(3*1)}{(2*1)-(3*3)}=\frac{10-3}{2-9} =\frac{7}{-7}=-1

Now, we can find the answer which is x= 2 and y= -1, we can replace these values in the equation to confirm the results are right, with the first equation:

  • 2x + 3y = 1
  • 2(2) + 3(-1)= 1
  • 4 - 3 = 1
  • 1 = 1

And, with the second equation:

  • 3x + y = 5​
  • 3(2) + (-1) = 5
  • 6 - 1 = 5
  • 5 = 5

 

4 0
3 years ago
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