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Lana71 [14]
3 years ago
10

(pls help i have this problem) The product of two positive consecutive integers is 132. find the integers.

Mathematics
1 answer:
masya89 [10]3 years ago
5 0
66+66 possibly .....the only thing i can come up with
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Find a . b if a=(2, -8) and b = (-1, 4).<br> a. -30<br> b. (1, -4)<br> c. -34<br> d. (-2, -32)
Luden [163]

Answer:

(2,-8)*(-1,4)

a*b

the answer is (-2,-32)

(2)*(-1)=(-2)

(-8)*(4)=(-32)

Step-by-step explanation:

hope this helps you :)

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4 years ago
Select all of the products of 2.74 • 7.5
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the ratio of the number of adults to the number of students at the prom has to be 1:10 lat year there were 477 more students tha
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3 years ago
Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ&gt;0
eduard
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2787701

_______________


\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\&#10;\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\&#10;\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}


Square both sides:

\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\&#10;\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\&#10;\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\&#10;\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}

\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\&#10;\mathsf{17\,sin^2\,\theta=1}\\\\&#10;\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\&#10;\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\&#10;\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}


Since \mathsf{sin\,\theta} is positive, you can discard the negative sign. So,

\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}


Substitute this value back into \mathsf{(i)} to find \mathsf{cos\,\theta:}

\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\&#10;\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}


I hope this helps. =)


Tags:   <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

7 0
3 years ago
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