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3 years ago

George found that he blinked 85 times in 15 minutes. At this rate, how many seconds passed during 51 blinks?

1 answer:

6 0

First convert the 15 minutes to seconds
60 sec-= 1 minute
900 sec = 15 minutes

ratio and proportion

85 blinks/900 sec= 51 blinks/ x seconds

cross multiply
85x=45900
divide by 85
x= 540 seconds
hope this helps

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We know the following about the numbers a, b and c:

labwork [276]

Step-by-step explanation:

(a + b)² = 9

(b + c)² = 25

(a + c)² = 81

Taking the square root:

a + b = ±3

b + c = ±5

a + c = ±9

By adding these three equations together and dividing both sides by 2, we get the value of a + b + c.

Possible combinations for a + b + c such that the sum is greater than or equal to 1 are:

a + b + c = (-3 + 5 + 9)/2 = 11/2

a + b + c = (3 − 5 + 9)/2 = 7/2

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3 years ago

(−5)3x7(yz)4 / (3)2x2y8z2

g100num [7]

Answer:

$\dfrac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}=\dfrac{-5x^5z^2}{2y^4}$

Step-by-step explanation:

Given:

The expression to simplify is given as:

$\frac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}$

In order to simplify this, we have to use the law of indices.

1. $(ab)^m=a^mb^m$

So, $(yz)^4=y^4z^4$

Substitute this value in the above expression. This gives,

$=\dfrac{(-5)3x^7y^4z^4}{(3)2x^2y^8z^2}\\\\\\=\dfrac{-15x^7y^4z^4}{6x^2y^8z^2}......(-5\times 3=15\ and\ 3\times 2=6)$

Now, we use another law of indices.

2. $\frac{a^m}{a^n}=a^{m-n}$

So, $\frac{x^7}{x^2}=x^{7-2}=x^5,\frac{y^4}{y^8}=y^{4-8}=y^{-4}, \frac{z^4}{z^2}=z^{4-2}=z^2$

Substitute these values in the above expression. This gives,

$=\frac{-15}{6}\times x^5\times y^{-4}\times z^2\\\\=\frac{-5x^5y^{-4}z^2}{2}$

Finally, we further simplify it using the law $a^{-m}=\frac{1}{a^m}$

So, $y^{-4}=\frac{1}{y^4}$

Therefore, the given expression is simplified as:

$\dfrac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}=\dfrac{-5x^5z^2}{2y^4}$

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