Matter is in a liquid state when its temperature is between its melting point and its boiling point. Suppose that some substance has a melting point of negative 35.84 degrees Upper C and a boiling point of 310.69 degrees Upper C . What is the range of temperatures in degrees Fahrenheit for which this substance is not in a liquid state? (Hint: Upper C equals five ninths left parenthesis Upper F minus 32 right parenthesis ) Express the range as an inequality. Let x represent the temperature in degrees Fahrenheit. What is the range of temperatures for which this substance is not in a liquid state? nothing (Type an inequality or a compound inequality. Simplify your answer. Use integers or decimals for any numbers in the expression. Round to three decimal places as needed.)

96.512 °F < x < 591.242 °F

Step-by-step explanation:

We are given;

Melting point = 35.84 °C

Boiling point = 310.69 °C

Now, we are given that formula to convert °C to °F is;

°C = (5/9)°F

Thus if the temperature in Fahrenheit is x, then;

Melting point is; 35.84 °C = (5/9)x°F - "32"

x = ((9 × 35.84)/5) + 32

x = 96.512 °F

Similarly, for Boiling point;

Boiling point is;

310.69 °C = (5/9)x°F - "32"

x = ((9 × 310.69)/5) + 32

x = 591.242 °F

Now, we are told that matter is in its liquid form when it is between melting and boiling point.

Thus, range of x in inequality form is;

96.512 °F < x < 591.242 °F

step-by-step explanation:

10.50 + taxes which added up to 11.50

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}