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Bogdan [553]
2 years ago
6

4% of 50 Percent of a quantity M

Mathematics
1 answer:
Ksivusya [100]2 years ago
6 0

Answer:

4% of 50 is 2

Step-by-step explanation:

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The main sail of a sailboat has the dimensions shown in the figure at the right. What is the height of the main sail?
levacccp [35]

Answer:

h = 16.1

Step-by-step explanation:

sin A /a = sin B / b = sin C/c   where A is angle and a is side opposite of angle

sin (49) / h = sin(41)/ 14       cross-multiply

h*sin(41) = 14 * sin(49)

h = (14*sin(49))/ sin(41)

h = 16.1

5 0
2 years ago
Vlad spent 20 minutes on his history homework and then completely solved x math problems that each took 2 minutes to complete. W
Veronika [31]
To solve you first need to pick out the most important information,
~Spent 20 minutes 
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Lets have y be the total time spent, and x the number of probs
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3 years ago
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Brut [27]
The answer is C I think
4 0
2 years ago
Point G is the centroid of the right △ABC with m∠C=90° and m∠B=30°. Find AG if CG=4 ft.
o-na [289]

Answer: \text{Length of AG=}\frac{2\sqrt{63}}{3}

Explanation:  

Please follow the diagram in attachment.  

As we know median from vertex C to hypotenuse is CM  

\therefore CM=\frac{1}{2}AB

We are given length of CG=4  

Median divide by centroid 2:1  

CG:GM=2:1  

Where, CG=4

\therefore GM=2 ft

Length of CM=4+2= 6 ft  

\therefore CM=\frac{1}{2}AB\Rightarrow AB=12

In \triangle ABC, \angle C=90^0

Using trigonometry ratio identities  

AC=AB\sin 30^0\Rightarrow AC=6 ft

BC=AB\cos 30^0\Rightarrow BC=6\sqrt{3} ft  

CN=\frac{1}{2}BC\Rightarrow CN=3\sqrt{3} ft

In \triangle CAN, \angle C=90^0  

Using pythagoreous theorem  

AN=\sqrt{6^2+(3\sqrt{3})^2\Rightarrow \sqrt{63}

Length of AG=2/3 AN

\text{Length of AG=}\frac{2\sqrt{63}}{3} ft


5 0
3 years ago
Andy filled four glasses with water at different temperatures as shown under each glass.
iragen [17]

Answer:

black

Step-by-step explanation:

because

8 0
3 years ago
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