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mestny [16]
4 years ago
12

Which expression is equivalent to

Mathematics
1 answer:
Damm [24]4 years ago
4 0

Answer:

\frac{(2a+1)^{2} }{50a}

Step-by-step explanation:

\frac{2a+1}{10a-5} * \frac{4a^{2} - 1}{10a}\\ = \frac{2a+1}{5(2a-1)} * \frac{(2a+1)(2a-1)}{10a}\\=\frac{(2a+1)^{2} }{50a}

1.The expression given is in division so when changing to multiplication the fraction is reversed.

2.Using algebra identity a^{2}  - b^{2}  = (a+b)(a-b).

3. Taking the common factor 5 out.

4. Finally cancelling the like terms from the numerator and denominator.

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2.14

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2 + 7/8 :)

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Dmitriy789 [7]
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3 years ago
A golf ball, thrown upwards, rises at a speed of v metres per second.
miss Akunina [59]

Answer:

162 metres

Step-by-step explanation:

Since h is proportional to the square of v, we know that their ratio must be constant, so v_1^2/h_1 = v_2^2/h_2 where v1 and v2 are velocities and h1 and h2 are their respective heights.

Since we are given that v = 10 and h = 8, we can set v1 = 10 and h1 = 8 and since we are trying to find the height for v = 45, we can set v2 = 45. Inputting these values into the equation and solving, we get

10^2/8 = 45^2/h2

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I hope this helps!

4 0
2 years ago
Emily has 12 fruits in her bowl. She has 3 apples, 5 bananas, 1 pear and 3 oranges. A fruit is selected at random. What is the p
blsea [12.9K]

Answer: The probability that the fruit is an orange or a pear is \dfrac{1}{3} .

Step-by-step explanation:

Given : Emily has 12 fruits in her bowl.

She has 3 apples, 5 bananas, 1 pear and 3 oranges.

A fruit is selected at random.

P(orange)  = \dfrac{\text{Number of oranges}}{\text{Total fruits}}

=\dfrac{3}{12}

P(pear)= \dfrac{\text{Number of pears}}{\text{Total fruits}}

=\dfrac{1}{12}

Since both events of selecting orange and pear are mutually exclusive , so

The probability that the fruit is an orange or a pear = P(orange) + P(pear)

=\dfrac{3}{12}+\dfrac{1}{12}=\dfrac{4}{12}=\dfrac{1}{3}

Therefore , the probability that the fruit is an orange or a pear is \dfrac{1}{3} .

5 0
4 years ago
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