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Harman [31]
2 years ago
11

Which equation can be used to determine the value of x?

Mathematics
2 answers:
lorasvet [3.4K]2 years ago
7 0

Answer:

(3x + 18) + 93 = 180

23 + X

Step-by-step explanation:

(3x + 18) + 93 = 180, its supplymentery

180 - 93 = 87

87 - 18 = 69

3x = 69

69/3 = 23

23 = X

Otrada [13]2 years ago
6 0

Answer:

Step-by-step explanation:

Remark

You have to assume nothing you should have to assume. You have to assume that 93 degrees on 3x + 18 sit on the same straight line. If it just looks that way, you are on shaky ground. You must be told. It's a good thing to keep in mind. Likely you'll get more questions like this. Just be aware you should not have to assume anything to make the problem work.

If the two angles both have one leg on what is marked as a straight line, then the sum of the two angles must = 180

Solution

The first one is not true. 3x + 18 is smaller than 90 degrees because 93 is more than 90 degrees. If A is true then the line is bent a bit.

B is not true. We just found one equation that doesn't work.

C is not true either. The result of trying to solve C will make x minus because 93>90 and 18 when subtracted  makes 3x even more minus.

D is true

(3x + 18) + 93 = 180               Subtract 93 from both sides.

(3x + 18) + 93-93 = 180 - 93   Combine

(3x + 18) = 87                           Remove the brackets

3x + 18 = 87                             Subtract 18

3x+18-18 = 87-18                

3x = 69                                    Divide by 3

3x/3 = 69/3

Answerx = 23

E is not true because D is true.

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Please solve this<br> (Rational numbers 8th grade)
pochemuha

                                        Question # 1

Answer:

\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=-\frac{3}{20}

Step-by-step explanation:

Given the expression

\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}

=-\frac{2}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}           ∵  \frac{-2}{3}\times \frac{3}{5}=-\frac{2}{5}

=-\frac{2}{5}+\frac{1}{4}                        ∵   \frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=\frac{1}{4}

\mathrm{Least\:Common\:Multiplier\:of\:}5,\:4:\quad 20

=-\frac{8}{20}+\frac{5}{20}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

=\frac{-8+5}{20}

\mathrm{Add/Subtract\:the\:numbers:}\:-8+5=-3

=\frac{-3}{20}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}

=-\frac{3}{20}

Therefore,

\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=-\frac{3}{20}

                                               Question # 2

Answer:

\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=-\frac{5}{28}

Step-by-step explanation:

Given

\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}

=-\frac{6}{35}-\frac{1}{6}\times \frac{3}{2}\times \frac{2}{5}\times \frac{1}{14}          ∵    \frac{2}{5}\times \frac{-3}{7}=-\frac{6}{35}

=-\frac{6}{35}-\frac{1}{140}         ∵   \frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=\frac{1}{140}

\mathrm{Least\:Common\:Multiplier\:of\:}35,\:140:\quad 140

=-\frac{24}{140}-\frac{1}{140}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

=\frac{-24-1}{140}

\mathrm{Subtract\:the\:numbers:}\:-24-1=-25

=\frac{-25}{140}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}

=-\frac{25}{140}

\mathrm{Cancel\:the\:common\:factor:}\:5

=-\frac{5}{28}

Therefore,

\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=-\frac{5}{28}

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