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wel
3 years ago
7

Find the value of x in a triangle with one side 11 and 1 angle being 28​

Mathematics
2 answers:
Kobotan [32]3 years ago
8 0

Answer:

c 12.5

Step-by-step explanation:

cos 28 = adjacent side/ hypotenuse

cos 28 = 11/x

Multiply each side by x

x cos 28 = 11/x *x

x cos 28 = 11

Divide each side by cos 28

x cos 28/ cos 28 =11 /cos 28

x = 11 /cos 28

x =12.45827056

To the nearest tenth

x = 12.5

Andre45 [30]3 years ago
4 0

Answer:

C: 12.5

Step-by-step explanation:

The sides x and 11 could be defined as the Adjacent angle and the Hypotenuse. This means that we will use the cos function to solve this.

First we can set up our equation

cos28=\frac{11}{x}

Next we can solve for x by multiplying by x and dividing by cos 28

x=\frac{11}{cos28}\\\\x=12.458\\\\x=12.5

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Answer:

Therefore the third container contains   \frac{735}{17}\% = 43.23 %  acid.

Step-by-step explanation:

Given that, One container is filled with the mixture that 25% acid. 55% acid is contain by second container.

Let the volume of first container be x cubic unit.

Since the volume of second container is 55% larger than the first.

Then the volume of the second container is

= (V\times \frac{100+55}{100})   cubic unit.

=\frac{155}{100}V  cubic unit.

The amount of acid in first container is

=V\times \frac{25}{100}  cubic unit.

=\frac{25}{100}V  cubic unit.

The amount of acid in second container is

=(\frac{155}{100}V \times \frac{55}{100}) cubic unit.

=\frac{8525}{10000}V cubic unit.

Total amount of acid =(\frac{25}{100}V+\frac{8525}{10000}V) cubic unit.

                                   =(\frac{2500+8525}{10000}V) cubic unit.

                                   =(\frac{11025}{10000}V)cubic unit.

Total volume of mixture =(V+\frac{155}{100}V) cubic unit.

                                       =\frac{100+155}{100}V cubic unit.

                                       =\frac{255}{100}V  cubic unit.

The amount of acid in the mixture is

=\frac{\textrm{The volume of acid}}{\textrm{The amount of mixture}}\times 100 \%

=\frac {\frac{11025}{10000}V}{\frac{255}{100}V}\times 100\%

=\frac{735}{17}\%

Therefore the third container contains  \frac{735}{17}\%  acid.

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