You want to isolate the x-term from the constant term, so you can subtract x/3 and add 10. This gives you
... 4/9x -10 -x/3 +10 > x/3 -12 -x/3 +10
... 1/9x > -2 . . . . . . collect terms
Now, you can multiply by 9 to see the condition on x.
... 9(1/9x) > -2(9)
... x > -18
On the x-y plane, the graph of this will be a dashed line at x=-18, and the half-plane to the right of that line will be shaded.
On a number line, there will be an open circle at x=-18, and the number line to the right of that circle will be marked (bold, colored, shaded, whatever).
Answer is provided in the image attached.
ANSWER
c. Domain: [0,18]
Range: [0,360]
EXPLANATION
The domain refers to the values of x for which the function is defined.
The given straight line graph is defined for x=0 to x=18.
Hence the domain is [0,18]
The range refers to the values of y for which x is defined.
The graph is defined for y=0 to y=360.
The range is [0,360]
Answer:
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
<u>Algebra I</u>
- Slope Formula:
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Choose any 2 points from the chart.</em>
Point (-6, 11)
Point (-2, 1)
<u>Step 2: Find slope </u><em><u>m</u></em>
- Substitute:
- Subtract/Add:
- Simplify:
Answer:
cement by 5 Kg
Step-by-step explanation:
add the parts of the ratio, 5 + 4 + 1 = 10 parts
Divide the quantity of concrete by 10 to find the value of one part of the ratio.
110Kg ÷ 10 = 11Kg ← value of 1 part of ratio , thus
5 parts = 5 × 11Kg = 55Kg ← cement required
4 parts = 4 × 11Kg = 44Kg ← sand required
1 part = 11Kg ← gravel required
He requires 55Kg of concrete but only has 50Kg.
He requires 44Kg of sand and has 55Kg
He requires 11Kg of gravel and has 15Kg
Thus he is 5Kg short of cement.
