Answer:
2
Step-by-step explanation:
i thought 7 is a base so just solve equation
2x+11=15
x=2
Answer:
23.83
Step-by-step explanation:
If the number next to it is above 5 then you round it up.
So the 2 becomes a 3 because the 6 to the right of it is above 5.
Hope this helps
Brainliest would be appreciated
Answer:
<h2>
![k = \frac{3D}{f}](https://tex.z-dn.net/?f=k%20%3D%20%20%5Cfrac%7B3D%7D%7Bf%7D%20)
</h2>
Step-by-step explanation:
<h3>
![D = \frac{1}{3} fk](https://tex.z-dn.net/?f=D%20%3D%20%20%5Cfrac%7B1%7D%7B3%7D%20fk)
</h3>
To solve for k first multiply both sides by 3
That's
<h3>
![3D = fk](https://tex.z-dn.net/?f=3D%20%3D%20fk)
</h3>
Next divide both sides by f in order to isolate k
We have
<h3>
![\frac{fk}{f} = \frac{3D}{f}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bfk%7D%7Bf%7D%20%20%3D%20%20%5Cfrac%7B3D%7D%7Bf%7D%20)
</h3>
Simplify
We have the final answer as
<h3>
![k = \frac{3D}{f}](https://tex.z-dn.net/?f=k%20%3D%20%20%5Cfrac%7B3D%7D%7Bf%7D%20)
</h3>
Hope this helps you
Prime numbers are numbers that can only be divided by itself and 1. The largest possible prime number that fits the scenario is 113
Let the prime numbers be p1 and p2, where p1 > p2; and the odd numbers be x1 and x2
So, we have:
![p_1 + p_2 + x_1 + x_2 = 128](https://tex.z-dn.net/?f=p_1%20%2B%20p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128)
The largest prime number less than 128 is 127.
If
, then
becomes
![127 + p_2 + x_1 + x_2 = 128](https://tex.z-dn.net/?f=127%20%2B%20p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128)
![p_2 + x_1 + x_2 = 128-127](https://tex.z-dn.net/?f=p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128-127)
![p_2 + x_1 + x_2 = 1](https://tex.z-dn.net/?f=p_2%20%2B%20x_1%20%2B%20x_2%20%3D%201)
<em>This is not possible, because three positive integers cannot add up to 1</em>
The next largest prime number is 113
If
, then
becomes
![113 + p_2 + x_1 + x_2 = 128](https://tex.z-dn.net/?f=113%20%2B%20p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128)
Collect like terms
![p_2 + x_1 + x_2 = 128-113](https://tex.z-dn.net/?f=p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128-113)
![p_2 + x_1 + x_2 = 15](https://tex.z-dn.net/?f=p_2%20%2B%20x_1%20%2B%20x_2%20%3D%2015)
Let ![p_2 = 3](https://tex.z-dn.net/?f=p_2%20%3D%203)
becomes
![3 + x_1 + x_2 = 15](https://tex.z-dn.net/?f=3%20%2B%20x_1%20%2B%20x_2%20%3D%2015)
Collect like terms
![x_1 + x_2 = 15-3](https://tex.z-dn.net/?f=x_1%20%2B%20x_2%20%3D%2015-3)
![x_1 + x_2 = 12](https://tex.z-dn.net/?f=x_1%20%2B%20x_2%20%3D%2012)
and
are odd numbers.
So, we have:
![x_1 = 5\\x_2 =7](https://tex.z-dn.net/?f=x_1%20%3D%205%5C%5Cx_2%20%3D7)
This is true because
![x_1 + x_2 = 12](https://tex.z-dn.net/?f=x_1%20%2B%20x_2%20%3D%2012)
![5 + 7 = 12](https://tex.z-dn.net/?f=5%20%2B%207%20%3D%2012)
<em>Hence, the largest of the possible primes is 113</em>
Read more about prime numbers at:
brainly.com/question/4184435
Answer:
13
Solution:
Using distance formula d=squarert[(x2-x1)^2+(y2-y1)^2] and plugging in the given values. You should get 13.