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4 years ago

How can I use fractions circles to find the sum of a mixed number?

Mathematics

1 answer:

Greeley [361]4 years ago

8 0

Answer:

Use fraction circles to add: 1% + 1% | Use fraction circles to model 1% and 1%. Whole and 1 Whole and 1 Whole and 1 eighth equals 3 wholes and 1 eighth. - Add the fractions and add the whole numbers separately. or: 0 Write each mixed number as an improper fraction, then add.

Step-by-step explanation:

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mia hired a moving company, it charged 500$ for service mia give them a 16% tip.How much did she tip the movers.

Novosadov [1.4K]

Answer:

$80

Step-by-step explanation:

First you take 500 and multiply by 16% or .16.

You should get 80.

A tip for the future is when you have a percentage just move the decimal place to the left twice.

ie. 50%=.50, 125%=1.25, 2%=.02

I hope this helps :)

6 0

3 years ago

Read 2 more answers

Find the value of 5x - 2y+ 3z when x = 7 , y = 11 and z = -1

Oksanka [162]

By replacing you get:

5*7-2*11+3*(-1) = 35-22-3= 10

so the answer is 10.

If you like my answers, feel free to follow me and ask me more maths problems. :)

7 0

3 years ago

Read 2 more answers

Please answer anyone you can will appreciate it

Pavel [41]

Well, for the first question, the answers will be (a) 13 per row, because 13*13=169, and (b) a square, because the dimensions will be 13x13.

8 0

3 years ago

Solve the following recurrence relation: <br> <img src="https://tex.z-dn.net/?f=A_%7Bn%7D%3Da_%7Bn-1%7D%2Bn%3B%20a_%7B1%7D%20%3D

-Dominant- [34]

By iteratively substituting, we have

$a_n = a_{n-1} + n$

$a_{n-1} = a_{n-2} + (n - 1) \implies a_n = a_{n-2} + n + (n - 1)$

$a_{n-2} = a_{n-3} + (n - 2) \implies a_n = a_{n-3} + n + (n - 1) + (n - 2)$

and the pattern continues down to the first term $a_1=0$ ,

$a_n = a_{n - (n - 1)} + n + (n - 1) + (n - 2) + \cdots + (n - (n - 2))$

$\implies a_n = a_1 + \displaystyle \sum_{k=0}^{n-2} (n - k)$

$\implies a_n = \displaystyle n \sum_{k=0}^{n-2} 1 - \sum_{k=0}^{n-2} k$

Recall the formulas

$\displaystyle \sum_{n=1}^N 1 = N$

$\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2$

It follows that

$a_n = n (n - 2) - \dfrac{(n-2)(n-1)}2$

$\implies a_n = \dfrac12 n^2 + \dfrac12 n - 1$

$\implies \boxed{a_n = \dfrac{(n+2)(n-1)}2}$

4 0

3 years ago

Find the coordinates of point X that lies along the directed line segment from Y(-8, 8) to T(-15, -13) and partitions the segmen

Nataly_w [17]

Answer:

C. (-13, -7)

Step-by-step explanation:

The location of a point O(x, y) that divides a line AB with location A $(x_1,y_1)$ and B $(x_2,y_2)$ in the ratio m:n is given by:

$x=\frac{m}{m+n} (x_2-x_1)+x_1\\\\y=\frac{m}{m+n} (y_2-y_1)+y_1$

Therefore the coordinates of point X That divides line segment from Y(-8, 8) to T(-15, -13) in the ratio 5:2 is:

$x=\frac{5}{5+2} (-15-(-8))+(-8)\\\\x=\frac{5}{7} (-15+8)-8=\frac{5}{7}(-7)-8=-5-8=-13 \\\\\\y=\frac{5}{5+2} (-13-8)+8\\\\y=\frac{5}{7} (-21)+8=5(-3)+8=-15+8=-7$

Therefore the coordinates of point X is at (-13, -7)

6 0

3 years ago