Answer:
Remove all perfect squares from inside the square root. ... I think it's about eighth or ninth grade. ... so if you have the cube root of the square root of (x-5) =2, you get ((x-5)^(1/2))^1/3 = 2, power to power requires multiplication, so (x-5)^1/6 = 2, ...
Missing: 176 xy
Answer:
∠ XZY ≈ 23.6°
Step-by-step explanation:
Using the sine ratio in the right triangle
sin Z = 
= 
= 
, thus
Z = 
() ≈ 23.6° ( to 1 dec. place )
The equation would be 40+15p=70+5p. You would subtract 5p from each side (15-5) and (5-5) and you'll have 40+10p=70. You would then subtract 40 from each side (70-40) and (40-40) to have 10p=30. You would divide by both of the sides by 10, you would end up with p=3. P is price training. The reason you divide and subtract on both sides is because you want the months to be on one side and the price training on the other side.
Step-by-step explanation:
Let 
Then 
or
This gives us or all integer multiples of 
Answer:
Step-by-step explanation:
We first need to define a couple of variables. Let s = the cost of 1 squash and z = the cost of 1 zucchini.
Now lets translate the words into algebra:
"The cost of 5 squash and 2 zucchini is $1.32" ===> 5s + 2z = 1.32
"Three squash and 1 zucchini cost $0.75" ===> 3s + z = 0.75
There are several ways to solve systems of equations. Let's use substitution. We can find what z equals in terms of s by manipulating the second equation:
3s + z = 0.75
-3s -3s
------------ -------------
z = -3s +0.75
Now lets substitute (-3s + 0.75) into the first equation for z, then solve for s:
5s + 2(-3s + 0.75) = 1.32
Can you handle it from here?
(Hint: Once you have solved for s, you can substitute that value back into either of the equations and solve for z.)
