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expeople1 [14]
3 years ago
9

Lamar hikes Trail B, which is 2 1/2. He stops every 5/6. How many stops does he make?

Mathematics
1 answer:
Dahasolnce [82]3 years ago
4 0

Answer:

2

Step-by-step explanation:

To solve this equation, you need to convert 2 1/2 into an improper fraction:

2 1/2 = 5/2

Then, you convert this into 6ths:

5/2 = 15/6

Next, you divide 15/6 by 5/6:

15/6 / 5/6 = 3

Because Jamal wouldn't make a stop at the end (it wouldn't be considered a stop because he is done), you subtract 1:

3-1=2

The answer is 2 (hope this helps, sorry if i got it wrong)

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-2.2 + 0.3z= 3 -0.5z -0.8z<br> what does z equal
wariber [46]
-2.2 + 0.3z= 3 -0.5z -0.8z \\ \\ 0.3z - 2.2 = 3 - 0.5z - 0.8z \\ \\ 0.3z - 2.2 = 3 - 1.3z \\ \\ 0.3z = 3 - 1.3z + 2.2 \\ \\ 0.3z = -1.3z + 5.2 \\ \\ 0.3z + 1.3z = 5.2 \\ \\ 1.6z = 5.2 \\ \\ z = 3.25 \\ \\

The final result is: z = 3.25.
3 0
3 years ago
Read 2 more answers
A bakery made 47 muffins that contained nuts and 3 muffins that did not contain nuts. What is the ratio of the number of muffins
gtnhenbr [62]

Answer:

Step-by-step explanation:

47= 3/44, so ratio is 3/44 for non-nuts muffins

Which mean, for every 44 muffins we would have 3 non-nuts muffins.

7 0
2 years ago
Which property is not used to simplify the following expression (x+2)×5-7=5 (x+2)-7
bazaltina [42]
Associative
** property I believe.
8 0
3 years ago
4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.
nikitadnepr [17]

Answer:

\sum_{n=0}^9cos(\frac{\pi n}{2})=1

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}

Step-by-step explanation:

\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))

=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})

=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1

2nd

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}

=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0

3th

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))

=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}

What we use?

We use that

e^{i\pi n}=cos(\pi n)+i sin(\pi n)

and

\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}

6 0
3 years ago
How to solve for 93,94, and 95
kvasek [131]

Answer:

93- not a triangle

94- right triangle

95- I dont know sorry!


3 0
3 years ago
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