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3 years ago

What is the length of BC? Round to the nearest tenth.

Mathematics

1 answer:

Kay [80]3 years ago

6 0

Answer:

BC ≈ 14.5 cm

Step-by-step explanation:

Using the sine ratio in the right triangle

sin65° = $\frac{opposite}{hypotenuse}$ = $\frac{BC}{AB}$ = $\frac{BC}{16}$ ( multiply both sides by 16 )

16 × sin65° = BC , then

BC ≈ 14.5 cm ( to the nearest tenth )

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How do you do multiplication in 7th grade mathimatics

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Answer:

Learn your timetables daily.

Step-by-step explanation:

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3 years ago

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A jar contains six blue marbles and five red marbles. Suppose you choose a marble at random, and do not replace it. Then you ch

TEA [102]

The jar has 6+5=11 marbles.

We have to find the probability of the following event:

1.We pick a marble from a jar that has 11 marbles in total, 5 of them are red

2.We pick a marble from a jar that has now 10 marbles in total, 4 of them are red (because in the previous step we picked a red marble and did not put it back in the jar)

The probability of the first event is:

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The probability of the second event is:

$P_2=\frac{4}{10}=\frac{2}{5}$

The probability of the both events to happen is:

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3 years ago

Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

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So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

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salantis [7]

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3 years ago

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tekilochka [14]

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3 years ago