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2 years ago

OK LAST ONE BUT PLEASE HELP 20 POINTS

Mathematics

1 answer:

I am Lyosha [343]2 years ago

5 0

Answer:

Solve for y, graph it, and then find where the lines intersect.You can take a table and put in all the x and y values. Then graph it. And see where it intersects. Because usually, these equations are lines

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Which statements are true about the trigonometric ratios of this triangle

rosijanka [135]

Answer:

Options (1), (2), (3) and (4)

Step-by-step explanation:

By applying the sine and cosine rules in the given triangle,

sinθ = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

cosθ = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

cos(30°) = $\frac{AC}{AB}$

= $\frac{11\sqrt{3} }{22}$

= $\frac{\sqrt{3}}{2}$

sin(30°) = $\frac{BC}{AB}$

= $\frac{11}{22}$

= $\frac{1}{2}$

cos(60°) = $\frac{BC}{AB}$

= $\frac{11}{22}$

= $\frac{1}{2}$

sin(60°) = = $\frac{AC}{AB}$

= $\frac{11\sqrt{3} }{22}$

= $\frac{\sqrt{3}}{2}$

Options (1), (2), (3) and (4) are the correct options.

3 0

2 years ago

Tju [1.3M]

Answer:

Step-by-step explanation:

Rationalize the denominator of b. So, multiply the numerator and denominator by $\sqrt{x}$

$b = \frac{(1-2\sqrt{x}) *\sqrt{x}}{\sqrt{x}*\sqrt{x} }=\frac{1*\sqrt{x} -2\sqrt{x} *\sqrt{x} }{\sqrt{x} *\sqrt{x} }\\\\=\frac{\sqrt{x} -2x}{x}\\$

Now, find a +b

$a +b = \frac{2x+\sqrt{x} }{x}+\frac{\sqrt{x} -2x}{x}\\\\=\frac{2x+\sqrt{x} +\sqrt{x} -2x}{x}$

Combine like terms

$= \frac{2x-2x+\sqrt{x} +\sqrt{x} }{x}\\\\=\frac{2\sqrt{x} }{x}$

Now find (a + b)²

(a +b)² = $(\frac{2\sqrt{x} }{x})^{2}$

$= \frac{2^{2}*(\sqrt{x} )^{2}}{x^{2}}\\\\= \frac{4* x}{x^{2}}\\\\= \frac{4}{x}$

Hint: $\sqrt{x} *\sqrt{x} =\sqrt{x*x}=x$

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2 years ago

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Answer:

Just substitute each number into the expression:

5(5) - 6(3) + 20(1/4)/4(3)(1/4) = 25 - 18 + 5/3 = 7 + 5/3 = 26/3 or 8 and 2/3.

Step-by-step explanation:

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