Answer:
x = 2
Step-by-step explanation:
Step 1: Write equation
6 - 7x = 7x - 9x - 4
Step 2: Solve for <em>x</em>
- Combine like terms: 6 - 7x = -2x - 4
- Add 7x to both sides: 6 = 5x - 4
- Add 4 to both sides: 10 = 5x
- Divide both sides by 5: 2 = x
- Rewrite x = 2
Step 3: Check
<em>Plug in x to verify it's a solution.</em>
6 - 7(2) = 7(2) - 9(2) - 4
6 - 14 = 14 - 18 - 4
-8 = -4 - 4
-8 = -8
Answer: Angle 2: 45°
Step-by-step explanation:
All triangles add up to 180, so all you would so is take both of those angles and subtract them from 180, so basically 180 - 135
Answer:
You could order 16 different kinds of pizza.
Step-by-step explanation:
You have those following toppings:
-Pepperoni
-Sausage
-Mushrooms
-Anchovies
The order is not important. For example, if you choose Sausage and Mushrooms toppings, it is the same as Mushrooms and Sausage. So we have a combination problem.
Combination formula:
A formula for the number of possible combinations of r objects from a set of n objects is:
![C_{(n,r)} = \frac{n!}{r!(n-r!}](https://tex.z-dn.net/?f=C_%7B%28n%2Cr%29%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Br%21%28n-r%21%7D)
How many different kinds of pizza could you order?
The total T is given by
![T = T_{0} + T_{1} + T_{2} + T_{3} + T_{4}](https://tex.z-dn.net/?f=T%20%3D%20T_%7B0%7D%20%2B%20T_%7B1%7D%20%2B%20T_%7B2%7D%20%2B%20T_%7B3%7D%20%2B%20T_%7B4%7D)
is the number of pizzas in which there are no toppings. So ![T_{0} = 1](https://tex.z-dn.net/?f=T_%7B0%7D%20%3D%201)
is the number of pizzas in which there are one topping
is a combination of 1 topping from a set of 4 toppings. So:
![T_{1} = \frac{4!}{1!(4-1)!} = 4](https://tex.z-dn.net/?f=T_%7B1%7D%20%3D%20%5Cfrac%7B4%21%7D%7B1%21%284-1%29%21%7D%20%3D%204)
is the number of pizzas in which there are two toppings
is a combination of 2 toppings from a set of 4 toppings. So:
![T_{2} = \frac{4!}{2!(4-2)!} = 6](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%20%5Cfrac%7B4%21%7D%7B2%21%284-2%29%21%7D%20%3D%206)
is the number of pizzas in which there are three toppings
is a combination of 3 toppings from a set of 4 toppings. So:
![T_{3} = \frac{4!}{3!(4-3)!} = 4](https://tex.z-dn.net/?f=T_%7B3%7D%20%3D%20%5Cfrac%7B4%21%7D%7B3%21%284-3%29%21%7D%20%3D%204)
is the number of pizzas in which there are four toppings. So ![T_{4} = 1](https://tex.z-dn.net/?f=T_%7B4%7D%20%3D%201)
Replacing it in T
![T = T_{0} + T_{1} + T_{2} + T_{3} + T_{4} = 1 + 4 + 6 + 4 + 1 = 16](https://tex.z-dn.net/?f=T%20%3D%20T_%7B0%7D%20%2B%20T_%7B1%7D%20%2B%20T_%7B2%7D%20%2B%20T_%7B3%7D%20%2B%20T_%7B4%7D%20%3D%201%20%2B%204%20%2B%206%20%2B%204%20%2B%201%20%3D%2016)
You could order 16 different kinds of pizza.
To solve for this, we need to figure out the sum of any two numbers on the dice that will become a pair factor for the listed answers.
What I mean by that is that we need to find the probability between how many two numbers will sum up to the answer.
3 can be 1 + 2, that's it. 3 has a probability of (1).
4 has a sum of 1 + 3, and 2 + 2, that's it. 4 has a probability of (2).
7 has a sum of 1 + 6, 2 + 5, and 3 + 4. 7 has a probability of (3).
11 has a sum of 5 + 6, that's it. 11 has a probability of (1).
The numbers in parenthesis are the probabilities.
7 has the highest number, so 7 has the highest probability.
Your answer is: 7.
I hope this helps!
Answer:
there both negative and gave the same y value I think I hope this helps