Use Pythagorean theorem.
[tex]|ON|^2+|MN|^2=|OM|^2\\\\x^2+x^2=8^2\\\\2x^2=64\ \ \ \ |divide\ both\ sides\ by\ 2\\\\x^2=32\to x=\sqrt{32}\\\\x=\sqrt{16\cdot2}\\\\x=\sqrt{16}\cdot\sqrt2\\\\\boxed{x=4\sqrt2}
Answer: |MN| = 4√2.
Answer:
a > c
Step-by-step explanation:
The Transitive Property of Inequality can be written as ...
If a > b and b > c, then a > c.
Based on the above, we can conclude from your premises that a > c.
For this case, the first thing we are going to do is assume that all the tests are worth the same.
Then, we define a variable:
x: score of Mona's last test
We write now the inequality that models the problem:
From here, we clear the value of x:
Answer:
the lowest grade that Mona can get for her last test so that her test average is 90 or more is:
x = 87
5x-3(0)=-2
5x=-2
X=-2 over 5 fractions mode
Answer:
2720
Step-by-step explanation:
because I used a calculator