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Umnica [9.8K]
3 years ago
12

Pls help me with dis question

Mathematics
2 answers:
Dima020 [189]3 years ago
6 0
I think the answer is 2/3, but I’m not too sure.

Can I pls have a brainliest??
mart [117]3 years ago
5 0

Answer:

3 1/15

Step-by-step explanation:

Product means multiply

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In an isosceles triangle, one of the equal interior angles is equal to the adjacent exterior
yarga [219]

Answer:

Step-by-step explanation:

can you show a picture?

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3 years ago
What is the vertex of a parabola defined by the equation <br> x = 5y2?
QveST [7]

Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0

3 0
4 years ago
If 4x is one factor of 4x^2-12x, what is the other factor?
Zepler [3.9K]
If you factor 4x²-12x, you can divide 4x from each term.

4x² ÷ 4x = x
-12x ÷ 4x = -3

Therefore, the other term becomes (x-3) because 4x(x-3) = 4x²-12x
4 0
3 years ago
PLSSS HELP ME THIS IS DUE TODAY THE QUESTION IS IN THE SCREENSHOT
expeople1 [14]

Answer:

b

Step-by-step explanation:

5 0
3 years ago
Find the exact value of the trigonometric expression.
Dmitrij [34]
\bf \textit{Half-Angle Identities}&#10;\\\\&#10;sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}&#10;\qquad &#10;cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\&#10;-------------------------------\\\\&#10;\cfrac{1}{12}\cdot 2\implies \cfrac{1}{6}\qquad therefore\qquad \cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}~thus~ \cfrac{\quad \frac{\pi }{6}\quad }{2}\implies \cfrac{\pi }{12}

\bf sec\left( \frac{\pi }{12} \right)\implies sec\left( \cfrac{\frac{\pi }{6}}{2} \right)\implies \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\impliedby \textit{now, let's do the bottom}&#10;\\\\\\&#10;cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}

\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}&#10;\\\\\\&#10;therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}


which simplifies thus far to 

\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}&#10;\\\\\\&#10;\cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{2^2-(\sqrt{3})^2}\implies \cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{1}

\bf 2\sqrt{2+\sqrt{3}}(2-\sqrt{3})\impliedby \stackrel{\textit{keep in mind that}}{(2-\sqrt{3})=(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}})}&#10;\\\\\\&#10;2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}&#10;\\\\\\&#10;2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}&#10;\\\\\\&#10;2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}&#10;\\\\\\&#10;2\sqrt{2-\sqrt{3}}
7 0
4 years ago
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