A bag of apples weighs 7 7/8 pounds. By weight, 1/15 of the apples are rotten. What is the weight of the good apples?

1 answer:

4 0

The good apples are 14/15 of the total apples. So multiply the total apples by the weight of good ones to find the solution. $\frac{63}{8}*\frac{14}{15} = \frac{294}{40} = 7\frac{7}{20}lbs$

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Plsssssss help<br>got 20 mins<br>the question is: The sin of angle DCB is

Elena L [17]

Answer:

i. <DCB = $53.13^{o}$

ii. Sin of <DCB = 0.8

Step-by-step explanation:

Let <DCB be represented by θ, so that;

Sin θ = $\frac{opposite}{hypotenuse}$

Thus from the given diagram, we have;

Sin θ = $\frac{4}{5}$

= 0.8

This implies that,

θ = $Sin^{-1}$ 0.8

= 53.1301

θ = $53.13^{o}$

Therefore, <DCB = $53.13^{o}$ .

So that,

Sin of <DCB = Sin $53.13^{o}$

= 0.8

Sin of <DCB = 0.8

5 0

2 years ago

The answer is red show step by step of the pro tem to get the answer

miskamm [114]

In order to rationalize the denominator of each expression, we need to multiply the expression by the same radical in the denominator, this way we can remove the radical from the denominator.

$\frac{5\sqrt{4}}{\sqrt{3}}(\cdot\sqrt{3})=\frac{5\sqrt{4}\sqrt{3}}{(\sqrt{3})^2}=\frac{5\cdot2\cdot\sqrt{3}}{3}=\frac{10\sqrt{3}}{3}$

10)

$-\frac{5}{3\sqrt{2}}(\cdot\sqrt{2})=-\frac{5\sqrt{2}}{3(\sqrt{2})^2}=-\frac{5\sqrt{2}}{3\cdot2}=-\frac{5\sqrt{2}}{6}$

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