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Alexxandr [17]
4 years ago
13

A wedding planner uses 72 ivy stems for 18 centerpieces. When she arrives at the venue, she realizes she will only need 16 cente

rpieces. How many ivy stems should she use so that the ratio of ivy stems to centerpieces stays the same? Type your answer as a number and a label. *
0 points
Mathematics
1 answer:
stepan [7]4 years ago
8 0
Wow that’s allot can u do it idk I know three is the answerrr
You might be interested in
can someone help me.Bob made 15 basketball shots in 6 seconds and then he ha 34 seconds left in does 34 seconds he made 85 baske
Anton [14]
You can add any number like 42 plus 42 gives you 84
3 0
3 years ago
Whitney needed to earn $400 to buy a ticket to a concert. If she earns $15 an hour babysitting, how many hours does she need to
MrMuchimi

Answer:

Shes needs to work 26.67 hours

Step-by-step explanation:

In order to find the amount of time she needs to work, we need to divide the total by the rate that she earns money at. She needs a total of $400 and she gets 15 every hour.

400/15=26.67

Shes needs to work 26.67 hours

4 0
3 years ago
Last question! Easy! :D Thank you all very much!
aalyn [17]
8 and 9 are the least possible integers for the answer I hope this is what you needed and it helped
8 0
4 years ago
Read 2 more answers
Area formula for 8c by 8c using A=bh
QveST [7]

Answer:

64 c^2

Step-by-step explanation:

If the base is 8c and the height is 8c

A = bh

   = (8c) * (8c)

    = 84 c^2

5 0
3 years ago
Please help! I've already answered part a, I don't understand what part b is asking.
Luba_88 [7]

Step-by-step explanation:

So, there is something known as a removable discontinuity, and it's essentially where you can define f(x) using the most simplified fraction, where you could normally not define f(x).

So we have the following equation:

f(x) = (\frac{x+5}{x+1}\div\frac{(x+3)(x-2)}{(x-4)(x+1)})-\frac{1}{x-2}

As you may know, we cannot divide a number by the value of zero. When the denominator is equal to zero, on the graph this will appear as a vertical asymptote, where x approaches the value that makes the denominator zero, but never actually reaches it.

If you look at each denominator, you can set them equal to zero to find the vertical asymptotes

x+1 = 0

x=-1

There should be a vertical asymptote at x=-1, since it would make two of the denominators equal to -1, but let's divide the two fractions first.

Original Equation

f(x) = (\frac{x+5}{x+1}\div\frac{(x+3)(x-2)}{(x-4)(x+1)})-\frac{1}{x-2}

Keep, change, flip

f(x) = (\frac{x+5}{x+1}*\frac{(x-4)(x+1)}{(x+3)(x-2)})-\frac{1}{x-2}

Multiply the two fractions

f(x) = (\frac{(x+5)(x-4)(x+1)}{(x+1)(x+3)(x-2)})-\frac{1}{x-2}

Notice how the x+1 is in the numerator and fraction? That means we can cancel it out!

f(x) = (\frac{(x+5)(x-4)}{(x+3)(x-2)})-\frac{1}{x-2}

In this simplified version of the fraction, we can technically define f(-1), but in the original version, since it's not defined there is a removable discontinuity at x=-1, meaning there is no vertical asymptote, but the function is still not defined at f(-1), and there will be a hole at that point.

4 0
2 years ago
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