Answer:
Step-by-step explanation:
First, we need to find the y-intercept. The y-intercept is the intersection of the point on the y-axis. We can clearly tell that the y-intercept is 1. The slope is definitely 0 because it shows a straight line.
Hence, the equation is y = 1.
Please check out my graph to support my reasoning.
Answer:
x+ 3/2y= 15
Step-by-step explanation:
8x+12y=120
To put the equation into standard form, the leading coefficient has to be 1. So, divide the whole equation by 8.
x+ 3/2y= 15
Answer:
a.)
b.)
c.) $
Step-by-step explanation:
a.) If she earns $7 'per' hour, this multiplication. After a week, we need to see how many hours she worked to earn $91, so this goes behind the equal sign. Let <em>h</em> be the unknown number of hours:
Solve for h. Divide both sides by 7:
Sylvia worked 13 hours at a rate of $7 an hour.
b.) She can work at most 15 hours a week. At 'most' can also mean less than or equal to. Let <em>h</em> be hours:
Sylvia can work less than or equal to 15 hours per week.
c.) If Sylvia can work 15 hours, plug in this value for <em>h</em> into an equation when she earns $7 per hour. Let the product of the hours and money equal <em>t</em> for total:
The most money Sylvia can earn in a week is $105.
<em>Finito.</em>
Answer:
f(1) = 9
Step-by-step explanation:
Evaluate f(x) = 4x+5 by replacing both instances of x with '1' as follows:
f(1) = 4(1) + 5 = 9
Thus, f(1) = 9
Answer:
See the argument below
Step-by-step explanation:
I will give the argument in symbolic form, using rules of inference.
First, let's conclude c.
(1)⇒a by simplification of conjunction
a⇒¬(¬a) by double negation
¬(¬a)∧(2)⇒¬(¬c) by Modus tollens
¬(¬c)⇒c by double negation
Now, the premise (5) is equivalent to ¬d∧¬h which is one of De Morgan's laws. From simplification, we conclude ¬h. We also concluded c before, then by adjunction, we conclude c∧¬h.
An alternative approach to De Morgan's law is the following:
By contradiction proof, assume h is true.
h⇒d∨h by addition
(5)∧(d∨h)⇒¬(d∨h)∧(d∨h), a contradiction. Hence we conclude ¬h.