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arlik [135]
3 years ago
12

Whats 2 5/3 reduced?

Mathematics
2 answers:
emmasim [6.3K]3 years ago
8 0
The answer will be 10/3 also I'll recommend downloading photo math it really helps!! It's free.
Sergio039 [100]3 years ago
8 0
11/3 is your answer
You might be interested in
Estimate the radius of the object
Anvisha [2.4K]

Answer:

r = 1.41 mm

Step-by-step explanation:

Given that,

The circumference of the flower, C = 8.9 mm

We need to find its radius. We know that,

Circumference, C=2\pi r

Where

r is the radius of the object

r=\dfrac{C}{2\pi}\\\\r=\dfrac{8.9}{2\pi}\\\\r=1.41\ mm

So, the radius of the object is equal to 1.41 mm.

7 0
3 years ago
Which choice is equivalent to the fraction below when x is an appropriate
garri49 [273]

The rationalized form of the expression will be -1+\sqrt{2}}

<h3>Rationalization of rational functions</h3>

Given the rational function

\frac{\sqrt{12}}{\sqrt{3}-3}\\

Multiplying by the conjugate of the denominator

=\frac{\sqrt{12}}{\sqrt{3}-3}\times \frac{\sqrt{3}+3}{\sqrt{3}+3}\\=\frac{\sqrt{12}(\sqrt{3}+3)}{3-9}\\ =\frac{(\sqrt{36}+3\sqrt{12})}{3-9}\\=\frac{6+6\sqrt{2}}{-6}\\ =

Simplifying further will give:

-1+\sqrt{2}}

Hence the rationalized form of the expression will be -1+\sqrt{2}}

Learn more on rationalization here: brainly.com/question/14261303

6 0
2 years ago
Debby new puppy weighed 10 1/8 pounds. After a month it had gained 6 1/3 pounds. What is the weight of the puppy after a month?
tatuchka [14]
The puppy after a month would weigh 16 11/24

I hope this helps

Have a happy holidays:)
7 0
3 years ago
What is the middle product of (-2x 1)(-3x - 5)? User: Factor 64a2 - 48a + 9.
Nikolay [14]
<span>What is the middle product of (-2x 1)(-3x - 5)
= </span><span>(-2x -1)(-3x - 5)
= 6x^2 + 3x + 10x + 5
= 6x^2 + 13x + 5

So the middle product is 13x.

</span><span>Factor 64a2 - 48a + 9.
</span><span>64a2 - 48a + 9
</span>(8a - 3) (8a - 3)
8 0
4 years ago
Can someone help? if A= H/n .290 h/300
Nesterboy [21]
I hope this helps you


0.290/300

0,00096666
8 0
4 years ago
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