Answer:
For the first one use the formula
y=x*2
For the second use
y=x*4
The third is
y=x*15
The fourth is
y=x*16
Step-by-step explanation:
To find the value of the y units, u have to multiply any x unit with the first y unit given but not 0
idk if it makes sense but i tried
good luck!!
The expression to solve is the
300-7 [4 (3+5)] + 3 to the 3rd power
3 to the third power means 3³, so
300-7 [4(3+5)]+3³
= 300 - 7 [4(8)] + 27
= 300 - 7[32] + 27
= 300 - 224 + 27
= 76 + 27
= 103
so, by solving this we get 103
A good place to start is to set 
to y. That would mean we are looking for 
to be an integer. Clearly, 
, because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since 
is a radical, it only outputs values from ![[0,\infty]](https://tex.z-dn.net/?f=%20%5B0%2C%5Cinfty%5D%20)
, which means y is on the closed interval: ![[0,120]](https://tex.z-dn.net/?f=%20%5B0%2C120%5D%20)
.
With that, we don't really have to consider y anymore, since we know the interval that is on.
Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval , which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:
Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.
N = -8
a a a a a a a a a a a a a a a a a a a a it’s right trust
:-):-):-):-):-)
Step-by-step explanation:
1) 40÷10
2) 40÷2
