Answer:
y=1X
Step-by-step explanation:
This is because the distance over the money made over the days worked is 2/2 which simplifies to 1 so that would be the one x. then the y intercept is found on the y axis where does the line cross the y axis? this would be at zero so you do not need to put that in the equation
Answer:
19.0681
Step-by-step explanation:
Given in the question that,
angle from ted to the dog = 60° with the ground
height of ted from the ground = 16ft
To find,
distance between dog and the door of ted's building
Considering the scenario make a right angle triangle:
<h3>By using pythagorus theorem:</h3>
Tan 40 = opposite / adjacent
Tan 40 = height / distance between dog and the door
Tan 40 = 16ft / x
x = 16 / tan40
x = 19.068057
x ≈ 19.0681 (nearest to thousand)
So, the dog need to walk 19.0681ft to reach the open door directly below Ted.
Answer:
360 combinations
Step-by-step explanation:
To calculate the number of different combinations of 2 different flavors, 1 topping, and 1 cone, we are going to use the rule of multiplication as:
<u> 6 </u>* <u> 5 </u> * <u> 4 </u>* <u> 3 </u>= 360
1st flavor 2nd flavor topping cone
Because first, we have 6 possible options for the flavor, then we only have 5 possible options for the 2nd flavor. Then, we have 4 options for the topping and finally, we have 3 options for the cone.
It means that there are 360 different combinations of two different flavors, one topping, and one cone are possible
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
The answer is true since the triangles are similar on all lengths and angles
