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Gwar [14]
3 years ago
13

What is the answer to the question?

Mathematics
1 answer:
marta [7]3 years ago
5 0

9514 1404 393

Answer:

  C)  √3

Step-by-step explanation:

The repeating decimals are both single digits, beginning at the decimal point. So, their equivalent is that digit divided by 9:

  0.\overline{4}=\dfrac{4}{9}\\\\0.\overline{9}=\dfrac{9}{9}=1

This means the given expression can be simplified to ...

  \dfrac{3\sqrt{0.\overline{4}}+\sqrt{0.\overline{9}}}{\sqrt{3}}=\dfrac{3\sqrt{\dfrac{4}{9}}+\sqrt{\dfrac{9}{9}}}{\sqrt{3}}=\dfrac{3\cdot\dfrac{2}{3}+1}{\sqrt{3}}=\dfrac{3}{\sqrt{3}}=\boxed{\sqrt{3}}

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Point A lies in plane B how many planes can be drawn perpendicular to plane B through point A
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Answer:

4

Step-by-step explanation:


8 0
3 years ago
Question 20 not yet answered marked out of 1.00 not flaggedflag question question text luna created a trash can in the shape of
Alex Ar [27]

Consider the triangle with sides a=1.6, b=2.3 and c=1.2 feet. You can use cosine theorem twice to find two unknown angles:

1.

b^2=a^2+c^2-2ac\cos \angle B,\\ (2.3)^2=(1.6)^2+(1.2)^2-2\cdot 1.6\cdot 1.2\cos \angle B,\\ 5.29=2.56+1.44-3.84\cos \angle B,\\ 3.84\cos \angle B=4-5.29=-1.29,\\\cos \angle B=-\dfrac{1.29}{3.84} =-0.336

Then you can conclude that ∠B is obtuse (because \cos \angle B) and m∠B=110°.

2.

c^2=a^2+b^2-2ab\cos \angle C,\\ (1.2)^2=(1.6)^2+(2.3)^2-2\cdot 1.6\cdot 2.3\cos \angle C,\\ 1.44=2.56+5.29-7.36\cos \angle C,\\ 7.36\cos \angle C=7.85-1.44=6.41,\\\cos \angle C=\dfrac{6.41}{7.36} =0.871.

Then you can conclude that ∠C is acute (because \cos \angle C>0) and m∠C=29°.

3. m∠A+m∠B+m∠C=180°, then m∠A=180°-110°-29°=41°.

Answer: m∠A=41° (∠A is opposite to the side a=1.6), m∠B=110° (∠B is opposite to the side b=2.3) and m∠C=29° (∠C is opposite to the side c=1.2)

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3 years ago
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