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BabaBlast [244]
10 months ago
6

How do I verify this identity? I know you should write sin2a as sin(a+a).

Mathematics
1 answer:
Kisachek [45]10 months ago
7 0

SOLUTION

Given the question in the image, the following are the solution steps to verify the identity

STEP 1: Write the given identity

\sin 2\alpha=2\sin \alpha\cos \alpha

STEP 2: Verify the identity

\begin{gathered} \sin 2\alpha=2\sin \alpha\cos \alpha \\ \text{Consider the left hand side of the above trigonometry identity.} \\ \text{That is, }\sin 2\alpha\text{.} \\ \text{ Rewrite }\sin 2\alpha\text{ as }\sin (\alpha+a) \\ \text{ It is known that }\sin (a+b)=\sin (a)\cos b+\cos (a)\sin (b) \\ U\sin g\text{ this statement above, we have;} \\ \sin (\alpha+a)=\sin a\cos \alpha+\cos a\sin \alpha \\ \text{It is known that }xy+yx=xy+xy=2\times xy=2xy \\ U\sin g\text{ this statement above, we have;} \\ \sin a\cos \alpha+\cos a\sin \alpha=\sin a\cos \alpha+\sin \alpha\cos \alpha=2\times\sin \alpha\cos \alpha=2\sin \alpha\cos \alpha \\ \text{Hence, }\sin 2\alpha=2\sin \alpha\cos \alpha \end{gathered}

The verification of the identity is as seen above.

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the density of a certain material is such that weighs 6900 grams for every 4.5 quarts of volume express this density in pounds p
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Answer:

d=101.4\ \text{pounds}/\text{feet}^3

Step-by-step explanation:

Given that,

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Volume of the material, V = 4.5 quarts

We need to find the density in pounds per cubic foot

We know that,

1 pound = 453.59 grams

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Also, 1 cubic foot = 29.92 quarts

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Density = mass/volume

Or

d=\dfrac{15.21\ \text{pounds}}{0.15\ \text{feet}^3}\\\\d=101.4\ \text{pounds}/\text{feet}^3

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3 years ago
Find the solutions of the quadratic equation 14x^2+9x+10=014x
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Answer:

Option B. x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

14x^{2}+9x+10=0

so

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substitute in the formula

x=\frac{-9(+/-)\sqrt{9^{2}-4(14)(10)}} {2(14)}

x=\frac{-9(+/-)\sqrt{-479}} {28}

Remember that

i=\sqrt{-1}

substitute

x=\frac{-9(+/-)\sqrt{479}i} {28}  

x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i

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