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2 years ago

Solve for X, and show the equation you would use to solve.

Mathematics

1 answer:

kompoz [17]2 years ago

7 0

Answer:

x = 20

Step-by-step explanation:

There are both consecutive interior angles and are supplementary.

The equation should be written like this:

5x + 50 + x + 10 = 180

Step 1: Simplify both sides of the equation.

5x + 50 + x + 10 = 180

(5x+x) + (50+10) = 180 (Combine Like Terms)

6x + 60 = 180

Step 2: Subtract 60 from both sides.

6x + 60 − 60 = 180 − 60

6x = 120

Step 3: Divide both sides by 6.

$\frac{6x}{6} = \frac{120}{6}$

x = 20

Hope it helps!!

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All else being equal, a study with which of the following error ranges would be the most reliable?

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3 years ago

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A farmer sells carrots and squash. A customer buys 1 pound of carrots and 1.5 ponds of squash for $7.00. Another customer buys 4

Otrada [13]

Answer:

$1.75

Step-by-step explanation:

1 pound of carrot and 1.5 pounds of squash cost $7.00.

1c + 1.5s = 7 ----------------- [ 1 ]

4.5 pounds of carrot and 1.75 pound of squash cost $14.00.

4.5c + 1.75s = 14 ----------------- [ 2 ]

[ 1 ] x 4.5:

4.5c + 6.75s = 31.5 ----------------- [ 1a ]

[1a] - [2]:

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8 0

3 years ago

What are 26 divide by 4,340

mart [117]

$\frac { 26 }{ 4,340 } \\ \\ =\frac { 26 }{ 4\cdot { 10 }^{ 3 }+3\cdot { 10 }^{ 2 }+4\cdot { 10 }^{ 1 } } \\ \\ =\frac { 26 }{ 10\left( 4\cdot { 10 }^{ 2 }+3\cdot { 10 }^{ 1 }+4 \right) } \\ \\ =\frac { 2\cdot 13 }{ 2\cdot 5\left( 4\cdot { 10 }^{ 2 }+3\cdot { 10 }^{ 1 }+4 \right) } \\ \\ =\frac { 13 }{ 5\left( 4\cdot { 10 }^{ 2 }+3\cdot { 10 }^{ 1 }+4 \right) }$

$\\ \\ =\frac { 13 }{ 5\left( 400+30+4 \right) } \\ \\ =\frac { 13 }{ 2,000+150+20 } \\ \\ =\frac { 13 }{ 2,170 }$

------

13/2170 = 0.006 (to 3 decimal places)

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3 years ago

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3 years ago

If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"

seraphim [82]

Answer:

$a + b + c = \pi \\ = > c= \pi - a - b \\ < = > \tan(c) = \tan(\pi - a - b) = -\tan(a + b)$

Step-by-step explanation:

we have:

$\tan(a) + \tan(b) + \tan(c) \\ = \tan(a) + \tan(b) - \tan(a + b) \\ = \tan( a) + \tan(b) - \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{ ( \tan(a) + \tan(b) ) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (1)$

we also have:

$\tan(a) \tan(b) \tan(c) \\ = - \tan(a) \tan(b) \tan(a + b) \\ = \frac{ -(\tan( a ) + \tan(b) ) \tan(a) \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{( \tan(a) + \tan(b)) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (2)$

from (1)(2) => proven

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2 years ago