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2 years ago

Find the value of the trigonometric ratio. sin A

Mathematics

1 answer:

romanna [79]2 years ago

8 0

Answer:

sin A = 4/5

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

sin A = opp / hyp

sin A = 24/ 30

Dividing the top and bottom by 6

sin A = 4/5

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Explain whether the following statement is true or false.

Lena [83]

Answer:

I believe it is the second one false

Step-by-step explanation:

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3 years ago

Read 2 more answers

Expand and simplify<br> (x – 3)(2x - 2)

vekshin1

Answer:

2x^2 - 8x + 6

Step-by-step explanation:

2x*x+(-3x*2x)+(x*(-2))+6=

2x^2-8x+6

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2 years ago

Thomas reads 2/9 of a book on Monday and 1/6 of it on Tuesday. He reads twice as many pages on Wednesday as on Tuesday. What fra

dezoksy [38]

1/18 is left. Thomas has 1/18 left of his book. How? What denominator can 9&6 relate to? 54? Sure, but it could be lower. How about 18? Ok. We have 18 because 9x2 and 6x3 both equal 18... 2/9 becomes 4/18 because 18 divided by 9 is 2. You multiply 2 with the two from 2/9. 1/6 becomes 3/18 because 18 divided by 6 is 3. You then multiply 3 with the one from 1/6. Put it together and TA-DA... Your answer is 1/18

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3 years ago

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$