Answer:
Let's define the high temperature as T.
We know that:
"four times T, was more than 2*T plus 66°C"
(i assume that the temperature is in °C)
We can write this inequality as:
4*T > 2*T + 66°C
Now we just need to solve this for T.
subtracting 2*T in both sides, we get:
4*T - 2*T > 2*T + 66°C - 2*T
2*T > 66°C
Now we can divide both sides by 2:
2*T/2 > 66°C/2
T > 33°C
So T was larger than 33°C
Notice that T = 33°C is not a solution of the inequality, then we should use the symbol ( for the set notation.
Then the range of possible temperatures is:
(33°C, ...)
Where we do not have an upper limit, so we could write this as:
(33°C, ∞°C)
(ignoring the fact that ∞°C is something impossible because it means infinite energy, but for the given problem it works)
Answer:
The answer is "y < 2x - 4".
Step-by-step explanation:
Write in <u>slope-intercept</u> form, the formula is "y = mx + b"
Hey there!
6(3.14)^6
= 6(3.14^6)
= 6(3.14 * 3.14 * 3.14 * 3.14 * 3.14 * 3.14)
= 6(9.8596 * 9.8596 * 9.8596)
= 6(97.21171216 * 9.8596)
= 6(5,750.811583)
≈ 5750.81158328
Therefore, the answer should be:
5,750.81158328
Good luck on your assignment & enjoy your day!
~Amphitrite1040:)
Answer:
A
C
i don't get the first slide
Step-by-step explanation:
Group and factor
undistribute then undistribute again
remember
ab+ac=a(b+c)
this is important
6d^4+4d^3-6d^2-4d
undistribute 2d
2d(3d^3+2d^2-3d-2)
group insides
2d[(3d^3+2d^2)+(-3d-2)]
undistribute
2d[(d^2)(3d+2)+(-1)(3d+2)]
undistribute the (3d+2) part
(2d)(d^2-1)(3d+2)
factor that difference of 2 perfect squares
(2d)(d-1)(d+1)(3d+2)
77.
group
(45z^3+20z^2)+(9z+4)
factor
(5z^2)(9z+4)+(1)(9z+4)
undistribuet (9z+4)
(5z^2+1)(9z+4)