First number --- (2a-1)
middle number ---- (2a)
third number -----(2a+1)
2a-1+2a+2a+1=250
6a=250
a=250/6= 125/3
2a=2*(125/3)= 250/3 ---- is not a whole number that is contradicts to what is given, so sum cannot be 250
so
Answer:
Step-by-step explanation:
1) Step 4: Marc has converted radical form to exponent form
Step 5: He has evaluated the value of the radical
3)
![x = \sqrt[3]{8}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%5B3%5D%7B8%7D)
Here, 3 is index and 8 is radicand.
Prime factorize 8 and for every 3 numbers we can take one number outside the radical
Solution:
![x = \sqrt[3]{2*2*2}\\\\x = 2](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%5B3%5D%7B2%2A2%2A2%7D%5C%5C%5C%5Cx%20%3D%202)
4) No,my equation does not have an extraneous solution.
My solution, satisfies the equation
Answer:
27
Step-by-step explanation:
it just is
Answer:
3½
Step-by-step explanation:
-3 - 1/6 + 6 + 2/3
3 + 4/6 - 1/6
3 + 3/6
3½
Answer:
En una bandeja, hay 375 pasteles. Si quitásemos 59 pasteles de otra bandeja, en ambas habría igual número de pasteles.
<h2>En cada bandeja, ¿Cuántos pasteles hay?</h2>