Math question...?

1 answer:

3 0

We have $\sqrt{x} \ \textless \ 16 ;$ ; where x >= 0;
Then, $( \sqrt{x}) ^{2} \ \textless \ 16^{2} ;$
Finally, 0 < x < 256;
The correct answer is A.

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What is the solution to the linear equation? 4b + 6 = 2 – b + 4

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Answer:

$\large\boxed{b=0}$

Step-by-step explanation:

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F=72

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------------

$\cos { \left( F \right) } =\frac { { e }^{ 2 }+{ g }^{ 2 }-{ f }^{ 2 } }{ 2eg }$

Therefore:

$\cos { \left( 72 \right) } =\frac { { e }^{ 2 }+{ 6 }^{ 2 }-{ f }^{ 2 } }{ 2\cdot e\cdot 6 } \\ \\ \cos { \left( 72 \right) } =\frac { { e }^{ 2 }+36-{ f }^{ 2 } }{ 12e }$

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But what is e?

E=76

G=32

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And:

$\frac { e }{ \sin { \left( E \right) } } =\frac { g }{ \sin { \left( G \right) } }$

Which means that:

$\frac { e }{ \sin { \left( 76 \right) } } =\frac { 6 }{ \sin { \left( 32 \right) } } \\ \\ \therefore \quad e=\frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } }$

If you take this value into account, you will discover that f is...

$f=\sqrt { \frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } \left( \frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } -12\cos { \left( 72 \right) } \right) +36 } \\ \\ \therefore \quad f=10.8\quad \left( 1\quad d.p \right)$

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