Subtract 153-63=90 and I am pretty sure that's what you have to do
Answer:
The false choices are A, C, and E
Step-by-step explanation:
Let's make an example:
p=1
q=2
s=sqrt(3)
t=sqrt(6)
Since pq=1*2=2, the answer is rational and A is false.
Since pt=1*sqrt(6)=sqrt(6), the answer is irrational and B is true.
Since p/q=1/2=0.5, the answer is rational and C is false.
Since st=sqrt(3)*sqrt(6)=sqrt(18), the answer is irrational and D is true.
Since s/t=sqrt(3)/sqrt(6)=sqrt(1/2), the answer is irrational and E is false.
You can write
1.5
⋅
10
5
=
15
⋅
10
4
15
⋅
10
4
−
8
⋅
10
4
=
(
15
−
8
)
⋅
10
4
=
7
⋅
10
4
Since the power is positive we move the dec.point 4 to the right:
=
70000
Hope it works
Let 1st integer = xLet 2nd integer = x + 1 We set up an equation. x(x + 1) = 195 x2 + x = 195 x2 + x - 195 = 0
We will use the quadratic formula: x = (-b ± √(b2 - 4ac) / (2a) x = (-1 ± √(1 - 4(-195))) / 2 x = (-1 ± √(781)) / 2 x = (-1 ± 27.95) / 2 x = 13.48x = -14.78
<span>We determine which value of x when substituted gives us a product of 195.</span> 13.48(14.48) = 195.19-14.48(-13.48) = 195.19 <span>The solution is 2 sets of two consecutive number</span> <span>Set 1</span> The 1st consecutive integer is 13.48The 2nd consecutive integer is 14.48
<span>Set 2</span> The 1st consecutive integer is -14.48The 2nd consecutive integer is -13.48Hopefully this helped, hard work lol :)
n, n + 1, n + 2, n + 3 - four consecutive integers
-142 - the sum
The equation:
n + (n + 1) + (n + 2) + (n + 3) = -142
n + n + 1 + n + 2 + n + 3 = -142
4n + 6 = -142 |subtract 6 from both sides
4n = -148 |divide both sides by 4
n = -37
n + 1 = -37 + 1 = -36
n + 2 = -37 + 2 = -35
n + 3 = -37 + 3 = -34
Answer: -37, -36, -35, -34.
