Need help , use pics ^

Mathematics

1 answer:

Rashid [163]3 years ago

8 0

Answer:

Step-by-step explanation:

Are there any other pictures of the actual problem?

This is proabbaly a transformation problem

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Please help me out please

natita [175]

Answer:

RC = 40

Step-by-step explanation:

Note that the circumcentre is equally distant from the triangle's 3 vertices.

That is : PC = RC = QC

Equate any pair and solve for x

Using RC = PC, then

5x - 15 = 3x + 7 ( subtract 3x from both sides )

2x - 15 = 7 ( add 15 to both sides )

2x = 22 ( divide both sides by 2 )

x = 11

Hence

RC = (5 × 11) - 15 = 55 - 15 = 40 units

5 0

3 years ago

How do u write 1.02 as a fraction and a percent

Brilliant_brown [7]

The anwser is one and two tenths

7 0

3 years ago

Read 2 more answers

Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )

Zina [86]

Answer:

$\purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

Step-by-step explanation:

$\cos( \alpha ) = \frac{ \sqrt{6} }{6} = \frac{1}{ \sqrt{6} } \\ \\ \therefore \: \sin( \alpha ) = \sqrt{1 - { \cos}^{2} ( \alpha ) } \\ \\ = \sqrt{1 - \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} } \\ \\ = \sqrt{1 - {\frac{1}{ {6} }}} \\ \\ = \sqrt{ {\frac{6 - 1}{ {6} }}} \\ \\ \red{\sin( \alpha ) = \sqrt{ { \frac{5}{ {6} }}} } \\ \\ \tan( \alpha ) = \frac{\sin( \alpha ) }{\cos( \alpha ) } = \sqrt{5} \\ \\ \sin( \beta ) = \frac{ \sqrt{2} }{4} \\ \\ \implies \: \cos( \beta ) = \sqrt{ \frac{7}{8} } \\ \\ \tan( \beta ) = \frac{\sin( \beta ) }{\cos( \beta ) } = \frac{1}{ \sqrt{7} } \\ \\ \tan( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha . \tan \beta} \\ \\ = \frac{ \sqrt{5} - \frac{1}{ \sqrt{7} } }{1 + \sqrt{5} . \frac{1}{ \sqrt{7} } } \\ \\ = \frac{ \sqrt{35} - 1 }{ \sqrt{7} + \sqrt{5} } \\ \\ \purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$