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Eva8 [605]
3 years ago
6

Is anyone wanna help me?

Mathematics
1 answer:
PtichkaEL [24]3 years ago
6 0

Answer:

See the answers bellow

Step-by-step explanation:

For 51:

Using the definition of funcion, given f(x) we know that different x MUST give us different images. If we have two different values of x that arrive to the same f(x) this is not a function. So, the pair (-4, 1) will lead to something that is not a funcion as this would imply that the image of -4 is 1, it is, f(-4)=1 but as we see in the table f(-4)=2. So, as the same x, -4, gives us tw different images, this is not a function.

For 52:

Here we select the three equations that include a y value that are 1, 3 and 4. The other values do not have a y value, so if we operate we will have the value of x equal to a number but not in relation to y.

For 53:

As he will spend $10 dollars on shipping, so he has $110 for buying bulbs. As every bulb costs $20 and he cannot buy parts of a bulb (this is saying you that the domain is in integers) he will, at maximum, buy 5 bulbs at a cost of $100, with $10 resting. He can not buy 6 bulbs and with this $10 is impossible to buy 0.5 bulbs. So, the domain is in integers from 1 <= n <= 5. Option 4.

For 54:

As the u values are integers from 8 to 12, having only 5 possible values, the domain of the function will also have only five integers values, With this we can eliminate options 1 and 2 as they are in real numbers. Option C is the set of values for u but not the domain of c(u). Finally, we have that 4 is correct, those are the values you have if you replace the integer values from 8 to 12 in c(u). Option 4.

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Lisa [10]
<span>The <u>correct answer</u> is:

$64.

Explanation<span>:

We will mulitiply the first two numbers first.

We will first write 1200 as a fraction by putting it over 1:
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We now multiply straight across: (1200*8)/(1*100) = 9600/100 = 96.

Now we will multiply this by the last fraction. Again we will make the whole number, 96, a fraction by placing it over 1:
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When we divide this out, we get 64.</span></span>
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