325 - [4(58 - 19) + (75 / 3)]
Divide:
325 - [4(58 - 19) + 25]
Distribute 4:
325 - [232 - 76 + 25]
Subtract:
325 - [156 + 25]
Add:
325 - [181]
Subtract:
144
To solve for x, we have to remember to isolate the variable.
For 1/2, we can make that 0.5, since their values are equivalent. Our equation:
Let's distribute the 0.5 first.
Now, let's simplify the right side of the equation. We have to distribute the negative to 3x and 1.
Then, we simplify the entire expression.
Our equation now:
Let's add 3x to the right and 3x to the left to simplify the -3x on the right side of the equation.
Let's do the same thing we did in Step 3 to 1.5. Subtract 1.5 on both sides of the equation.
Finally, we divide both sides by 6 to isolate x.
If we take the value of t to be 15 and then we perform the calculations of 2 times t divided by 3, we can express it all by two equations substituting the value given for t like so:
t = 15
2t/3 = 2(15)/3 = 30/3 = 10
Answer:
Ans A). The graph is shown.
Ans B). 18.3333 C temperature when F is 65 temperature
Ans C). 32 F when the line crosses the horizontal axis
Ans D). Slope of line C= is
Step-by-step explanation:
Given equation is C=
Ans A).
For the table,
Take the four value of F as 32,41,50,59.
For F = 32.
The value of C is
C=
C=
C=0.
For F = 41.
The value of C is
C=
C=
C=05
For F = 50.
The value of C is
C=
C=
C=10
For F = 59.
The value of C is
C=
C=
C=15
<em>Note: The figure shows a graph of given equation with points.</em>
Ans B). Estimate temperature in C when the temperature in F is 65
For F = 65.
The value of C is
C=
C=
<em>C=18.333333.</em>
Ans C). At what temperature, graph lien cross the horizontal axis
When the line crosses the horizontal axis, C=0
Therefore,
C=
0=
0=
F=32 Temperature.
Ans D). Slope of the line C=
The slope of line is given by s=
Take points from the table of answer A.
let (32,0) and (41,5) using for slope.
s=
s=
s=
Slope of line C= is