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2 years ago

Ismail tried to prove that sin(θ) = cos(90°− θ) using the following diagram. His proof is not correct.

Mathematics

1 answer:

Anton [14]2 years ago

5 0

Answer:

C. Ismail used the wrong sides in his ratio for cos (90 - θ)

Step-by-step explanation:

All steps correct apart from:

cos (90° - θ) = AC/BC

It should be

cos (90° - θ) = AB/BC

Correct choice is C

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If α and β are the zeros of the quadratic polynomial f(x) = 3x2–4x + 5, find a polynomialwhose zeros are 2α + 3β and 3α + 2β.

Lelechka [254]

Answer:

$\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }$

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

$f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)$

So we can say that

$a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}$

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

$(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab$

and we can notice that

$a^2+b^2=(a+b)^2-2ab$ so

$(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab$

it comes

$x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}$

multiply by 3

$3x^2-20x+2*16+5=3x^2-20x+37$

4 0

2 years ago

Which of the following is true? (

stiks02 [169]

Answer:

B. |-6|<|-7|

Step-by-step explanation:

|-6| = 6

|-7| = 7

and 6 <7 so the answer is B.

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2 years ago

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KATRIN_1 [288]

It would be in the top left corner

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This is what the graph would look like. Hope this helps:)

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2 years ago

How many lines of symmetry does the figure have? A. 0 B. 1 C. 2

balu736 [363]

Answer:

The answer is A, so zero.

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2 years ago