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Vanyuwa [196]
3 years ago
10

Select equivalent or nonequivalent for each pair of expressions

Mathematics
1 answer:
wel3 years ago
6 0

Answer:

i think the answer is a as well <3

have a great day!

also wdym by i smell pennies??

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How is 3/5 and 8/10 Are equivalent fractions?
olya-2409 [2.1K]
Hi!

These are not <span>equivalent fractions because 8/10 reduces to 4/5, not 3/5, therefore, they are not </span>equivalent.
4 0
3 years ago
Rosa earns $120 per week tutoring math. Each week, she puts $36 from her paycheck in her bank account to save for college. Rosa
natita [175]

Answer:

36/120 x 100

= 30%

..............

3 0
3 years ago
What is 9 5/9-6 5/6 I seem to not get the right answer every time I try
gulaghasi [49]
If you would like to solve 9 5/9 - 6 5/6, you can calculate this using the following steps:

9 5/9 - 6 5/6 = 86/9 - 41/6 = 516/54 - 369/54 = (516 - 369) / 54 = 147/54 = 49/18 = 2 13/18

The correct result would be 2 13/18.
6 0
3 years ago
Read 2 more answers
Find the sum. Write your answer in simplest form. - 4 1/6 + 5 4/7 ​
Kitty [74]

Answer:

1 17/42.

Step-by-step explanation:

Find the common denominator, which is 42.

So change -4 1/6 to -4 7/42.

And change 5 4/7 to 5 24/52.

So add -4 7/42 + 5 24/52 and you'll get: 1 17/42.

(The decimal form of this is 1.404762)

7 0
3 years ago
Read 2 more answers
Find the magnitude of the vector &lt;−4, −6&gt;.<br><br> 8<br> 7.2<br> 10<br> 6.4
PolarNik [594]

The magnitude of the vectors is 7.2

The magnitude of a vector is resolved by using a formula

<h3>Magnitude of a Vector</h3>

This is the displacement between two vectors and can be calculated as

x^2 = y^2 + z^2\\x = \sqrt{y^2 + z^2} \\

where x is the magnitude of displacement between y and z.

In this question, the vectors are

  • -4
  • -6

Let's find the magnitude of this vector

x = \sqrt{(-4)^2+(-6)^2}\\x = \sqrt{16+36}\\x = \sqrt{52}\\x = 7.2

The magnitude of the vectors is 7.2

Learn more on magnitude of vectors here;

brainly.com/question/3184914

brainly.com/question/8536161

4 0
2 years ago
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