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2 years ago

Find each value of x, determine whether it is a solution to 1-3x <-26

Mathematics

1 answer:

Aleks04 [339]2 years ago

8 0

<h3>Answer:</h3>

$\large\boxed{x>9}$

<h3>Step-by-step explanation:</h3>

In this question, we're trying to solve for x.

What this means is that we're trying to solve the inequality to find what x is.

The inequality we are solving is:

$1-3x$

We will get x by itself and solve after that.

Solve:

$1-3x$

When you're done solving, you should get 9.

This means that x > 9

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>

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Why does 20x700=14000 please give details

Butoxors [25]

Answer:

When multiplying zeros, you first remove the zeros.

So,

2 x 7 = 14

Then you add the zeros to the answer.

We had three zeros

So we add three zeros to 14

14,000

8 0

2 years ago

Find f.<br><br> A) 7.4<br><br> B) 8.2<br><br> C) 10.5<br><br> D) 11.1

tatyana61 [14]

F=72

g=6

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

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

Therefore:

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

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

But what is e?

E=76

G=32

g=6

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...

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

So I would have to say that the answer is approximately (c).

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Answer:

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