In London today, four times the high temperature was more than twice the high temperature plus sixty-six. In interval form, what are the possible temperatures

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)

9.55 degrees.

step-by-step explanation:

the given angle is

1 cycle is equal to 2π radians. so 4π will be equal to two cycles.

5.166667π-4π = 1.166667π

hence, this angle goes a little more than negative x axis.

1.166667π-π= 0.166667π

hence, the reference angle is 0.166667π radians.

1 radian = 57.2958 degrees

now, in degrees, this should be equals to 9.55 degrees.

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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