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r-ruslan [8.4K]
3 years ago
15

Which equation is a function?

Mathematics
1 answer:
-Dominant- [34]3 years ago
3 0

the answer is B -- no matter the x factor y should be single

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A bag contains 2 red marbles, 3 blue marbles, 4 white marbles, and 1 yellow marble. If Jamal picks a marble at random, notes the
statuscvo [17]

Answer:

About 99 times.

Step-by-step explanation:

There are 3 out of 10 blue marbles in the bag, or 0.3%. 99 is 0.3% of 330.

4 0
3 years ago
HELP PLEASE
Maurinko [17]
\bf \cfrac{x^2-x-42}{x-6}\implies \cfrac{(x+7)\underline{(x-6)}}{\underline{x-6}}\implies x+7

so.. the left-hand-side does indeed simplify to x+7, so the equation does check out.

however, notice something, for the equation of x+7, when x = 6, we get (6) + 7 which is 13.

BUT for the rational, we get    \bf \cfrac{x^2-x-42}{x-6}\qquad \boxed{x=6}\implies \cfrac{x^2-x-42}{\boxed{6}-6}\implies \stackrel{und efined}{\cfrac{x^2-x-42}{0}}

so, even though the siimplification is correct, the rational or original expression is constrained in its domain.
6 0
3 years ago
Given: h(x)=-4x and g(x) = 3x +1<br> Find:<br> (nog)
pashok25 [27]

Answer:

Step-by-step explanation:

h(g(x))

g(x)= 3x + 1

h(3x + 1) = 4(3x + 1) = 12x + 4

4 0
4 years ago
Suppose a box contains 3 defective light bulbs and 12 good bulbs. two bulbs are chosen from the box without replacement. find th
Mademuasel [1]
The probability that the bulb is good is 12/15.
Since we have taken 1 item out the remaining total is now 14, so the probability of getting a defective bulb is now 3/14.
Now you multiply the probabilities together to get (12/15)(3/14)=(4/5)(3/14)=12/70=6/35
7 0
4 years ago
Please help asap! this isn’t too hard but i am still confused.
juin [17]

Answer:

The consecutive even numbers are 28, 30, 32, 34

Step-by-step explanation:

let the four consecutive even numbers be: b d f h,

such that:

b=b, d=b+2, f=b+4, h=b+6.

Therefore, 2(b+b+2)=3(b+6)+14

b=28,

d=(b+2)=30

f=(b+4)=32

h=(b+6)=34

7 0
3 years ago
Read 2 more answers
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