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monitta
3 years ago
6

Help me Graph this Linear function 3y + 4y = 12

Mathematics
1 answer:
ivanzaharov [21]3 years ago
4 0
Here's the graph:

Put a straight horizontal line when the y value is equal to 7/12
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What is 5 3/5 - 4 1/10
Mariulka [41]

Answer:

3/2

Step-by-step explanation:

5 0
2 years ago
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Jayla ate 75% of the jelly beans in the bag. If she ate 24 jelly beans how many were in the bag?
maksim [4K]

Answer:

32 jelly beans were in the bag

Step-by-step explanation:

We know 75% is 3/4 and that she ate 24, so divide 24 by 3 then multiply it by 4:

24/3= 8

8 x 4 = 32

7 0
3 years ago
Consider the Polynomial P(x)=x^3-5x^2-x+5. is (x-5) a factor?
Alborosie

Answer:

Option B, Yes, the remainder is 0, so x-5 is a factor of P(x)

Step-by-step explanation:

<u>Step 1:  Factor</u>

p(x) = x^3 - 5x^2 - x + 5

<em>p(x) = (x - 5)(x + 1)(x - 1)</em>

<em />

<em>Yes, the remainder is 0 so, x-5 is a factor of p(x)</em>

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Answer:  Option B, Yes, the remainder is 0, so x-5 is a factor of P(x)

8 0
3 years ago
What are the points of discontinutity y=(x-3)/x^2-12x+27
madreJ [45]

Answer:

(3, -\frac{1}{6})

Step-by-step explanation:

We can rewrite the equation as

y = \frac{x - 3}{(x - 3)(x - 9)}

Notice that we have x - 3 in both the numerator and the denominator, so it looks like we can divide it out. However, what if x - 3 is 0? Then we would have y = \frac{0}{0 \times (x - 9)} = \frac{0}{0}, which is undefined. So although it looks like the numerator and denominator can be simplified, the resulting function we would get from simplification would not have the same behavior as this one (since such a function would be defined for x = 3, but this one is not).

A point of discontinuity refers to a particular point which is included in the simplified function, but which is not included in the original one. In this case, the point which is not included in the unsimplified function is at x = 3. In the simplified version of the function, if we plug in x = 3, we get

y = \frac{1}{((3) - 9)} = -\frac{1}{6}

So the point (3, -\frac{1}{6}) is our only point of discontinuity.

It's also important to distinguish between specific points of discontinuity and vertical asymptotes. This function also has a vertical asymptote at x = 9 (since it causes the denominator to be 0), but the difference in behavior is that in the case of the asymptote, only the denominator becomes 0 for a specific value of x

5 0
3 years ago
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Marcy is concerned that her findings may be due to an extraneous uncontrolled variable and not her treatment. Marcy is most conc
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Answer:

c

Step-by-step explanation:

because it's her extraneous uncontrolled variable

4 0
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