Answer:
15 units
Step-by-step explanation:
I just took this geometry test with the same question. Its 15
well, we know the ceiling is 6+2/3 high, and Eduardo has 4+1/2 yards only, how much more does he need, well, is simply their difference, let's firstly convert the mixed fractions to improper fractions and then subtract.
![\stackrel{mixed}{6\frac{2}{3}}\implies \cfrac{6\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{20}{3}} ~\hfill \stackrel{mixed}{4\frac{1}{2}}\implies \cfrac{4\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{9}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{20}{3}-\cfrac{9}{2}\implies \stackrel{using ~~\stackrel{LCD}{6}}{\cfrac{(2\cdot 20)-(3\cdot 9)}{6}}\implies \cfrac{40-27}{6}\implies \cfrac{13}{6}\implies\blacktriangleright 2\frac{1}{6} \blacktriangleleft](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B6%5Cfrac%7B2%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B6%5Ccdot%203%2B2%7D%7B3%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B20%7D%7B3%7D%7D%20~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B4%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B9%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B20%7D%7B3%7D-%5Ccfrac%7B9%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Busing%20~~%5Cstackrel%7BLCD%7D%7B6%7D%7D%7B%5Ccfrac%7B%282%5Ccdot%2020%29-%283%5Ccdot%209%29%7D%7B6%7D%7D%5Cimplies%20%5Ccfrac%7B40-27%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B13%7D%7B6%7D%5Cimplies%5Cblacktriangleright%202%5Cfrac%7B1%7D%7B6%7D%20%5Cblacktriangleleft)
For
the first question; we have a problem dealing with translations.
Our goal is to translate
the word problem into an algebraic one. So our numbers are
60, 2, and 5. Our variable, or the letter holding the unknown
number's place
value, is
N. <span>
</span>The
problem states that 60 is MORE THAN the PRODUCT of 5 and N. Focus
on the words I bolded and capitalized. These are your key words that
tells you what operations
you need to use. Note that 'product' signals the operation of
multiplication (Since the answer you get from multiplying numbers is
called the 'product'.) and
'more than' signals the operation of addition (Because you are left
off with more than you originally had before adding). <span>
</span>Now
that we know that our algebraic expression contains multiplication,
addition, letter N and the numbers 2 and 5; we can start plugging the
numbers in.<span>
</span>60
= 2 (more than) + (product of) 5N. The answer for the first
attachment is
B. <span>
</span>_____________________________________________________________<span>
</span>All
we need to do for the AB problem (Question 25 on your material) is
solve AB.<span>
</span>Note
1: When a problem has no operation separating two letters or numbers,
this means multiplication.<span>
</span>Note
2: A and B are variables. Variables are letters that stand for and
hold the place of our unknown quantities. However, the values are not
unknown anymore given the fact that the question tells us that A= 42
and B = 2. Thus we substitute A with 42 and B with 2, then work from
there. 42 * 2 =84. Answer = A.
____________________________________________________________________
And
finally, for question 26, we use PEMDAS. P = parenthesis (), E =
exponent x^2, M/D = x or /, A/S means + or subtraction.
According
to PEMDAS, we first solve what's parenthesis, which is 77 – 32.
Using mental math, 77 – 32 = 45. ( 7 – 3 = 4 and 7 – 2 = 5.). 5/9
* 45 is what we have left. 5 divided by 9 =0.5555555555555556.
0.5555555555555556 * 45 = 25. Answer is D.
Answer:
100
Step-by-step explanation:
Mixed candy question... Skittles jar... to be filled with Jelly beans.
Let's first calculate the volume of the jar. We'll assume it's a regular cylindrical prism jar, unlike the one on the photo which is narrower on top.
V = π * r² * h = π * (3.5)² * 11.5 = 140.875 π = 442.6 cubic cm
Now, we don't have the precise measurement of a jelly bean, but we know it's roughly 2-3 cubic cm. The precision isn't needed to answer this question, just to have a rough idea... it's no 300 cu cm per jelly bean.
So, let's assume a 3 cu cm per jelly bean (2 cu cm wouldn't the final answer)....
442.6 / 3 = 147.5 jelly beans, approximately.
So, can they fit 100,000? No
Can we fit 10,000 in there? No
Can we fit 100? Yes.
Can we fit 1? Certainly
The most reasonable lower-limit would then be 100.
