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3 years ago

Please correct my work and show your work if I MESSED up please

Mathematics

1 answer:

sertanlavr [38]3 years ago

4 0

Hello from MrBillDoesMath!

Answer:

Your answer is correct. (You have earned the unofficial title of Genius, Jr. );

Discussion:

Whew! This is a long problem. Let me see what I get and compare it to your answer. First each of the 3 terms contains the sqrt(yz) so let's factor that terms out. You equations equals

sqrt(yz) ( sqrt(108) + 3 sqrt(98) + 2 sqrt(75) ) (A)

Hmm.... let's look at the prime factorizations....

108 = 3 * 36 = 3 * (4*9) = 3 * 2^2* 3^2

98 = 2 * 49 = 2 * 7^2

75 = 3 * 25 = 3 * 5^2

So (A) =

sqrt(yz) ( 6 sqrt(3) + 3*7 sqrt(2) + 2*5 sqrt(3) ) =

sqrt(yz) ( ( 6 + 2*5) sqrt(3) + 21 sqrt(2) ) =

sqrt(yz) ( (6 + 10) sqrt(3) + 21 sqrt(2)) =

sqrt(yz) ( 16 sqrt(3) + 21 sqrt(2) )

Our answers agree!

Thank you,

MrB

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a car travels along a straight road to the east for 150 meters in 4 seconds and west for 50 meters in one second

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2 years ago

Scott and Letitia are brother and sister. After dinner, they have to do the dishes, with one washing and the other drying. They

ehidna [41]

Answer:

The probability that Scott will wash is 2.5

Step-by-step explanation:

Given

Let the events be: P = Purple and G = Green

$P = 2$

$G = 3$

Required

The probability of Scott washing the dishes

If Scott washes the dishes, then it means he picks two spoons of the same color handle.

So, we have to calculate the probability of picking the same handle. i.e.

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

This gives:

$P(G_1\ and\ G_2) = P(G_1) * P(G_2)$

$P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}$

$P(G_1\ and\ G_2) = \frac{3}{10}$

$P(P_1\ and\ P_2) = P(P_1) * P(P_2)$

$P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}$

$P(P_1\ and\ P_2) = \frac{1}{10}$

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

So, we have:

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

$P(Same) = \frac{3}{10} + \frac{1}{10}$

$P(Same) = \frac{3+1}{10}$

$P(Same) = \frac{4}{10}$

$P(Same) = \frac{2}{5}$

8 0

2 years ago

A bucket holds b liters of water. But a jar holds 8 fewer liters than the bucket. How many times can the bucket hold than the ja

kykrilka [37]

Answer:

b / (b - 8)

Step-by-step explanation:

Given that :

Litres of water held by bucket = b

Jar holds 8 litres fewer than bucket; hence, litres held by jar = b - 8

The number of times bucket can hold than jar is :

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2 years ago

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DaniilM [7]

Answer:

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Step-by-step explanation:

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Multiply.

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Answer:

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