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stira [4]
3 years ago
5

PLEASE HELP how many solutions does this equation have? 12+2x-8=7x+5-5x

Mathematics
1 answer:
slava [35]3 years ago
7 0

<u>Answer:</u>

no solution

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

STEP 1: <em>Since x is on the right side of the equation, switch the sides so it is on the left side of the equation.</em>

<em />7x+5-5x=12+2x-8<em />

STEP 2: <em>Subtract 5x from 7x.</em>

<em />2x+5=12+2x-8<em />

STEP 3: <em>Subtract 8 from 12.</em>

<em />2x+5=2x+4<em />

STEP 4: <em>Move all terms containing x to the left side of the equation.</em>

<em />5=4<em />

STEP 5:<em> Since 5≠4, there are no solutions.</em>

No solution

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Write the equation of a line parallel to y=7x+8 and goes through the point (-1,1).
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1. In ABC, C is a right angle and BC = 11. If B = 30°, find AC. (1 point)
faltersainse [42]

Answer:

AC=\frac{11\sqrt{3}}{3}

Step-by-step explanation:

Given that triangle ABC is a right angle triangle. Where angle C is a right angle. Also we have been given that BC = 11, B = 30°. Now we need to find the value of AC.

Apply formula:

\tan\left(\theta\right)=\frac{opposite}{adjacent}

\tan\left(B\right)=\frac{AC}{BC}

\tan\left(30^o\right)=\frac{AC}{11}

\frac{1}{\sqrt{3}}=\frac{AC}{11}

\frac{11}{\sqrt{3}}=AC

AC=\frac{11}{\sqrt{3}}

or

AC=\frac{11}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}

or

AC=\frac{11\sqrt{3}}{3}

Hence final answer is AC=\frac{11\sqrt{3}}{3}.

7 0
3 years ago
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Answer:

x              y

9             18

5             10

1              2

Step-by-step explanation:

Please let me know if you want me to add an explanation as to why this is the answer. I can definitely do that, I just wouldn’t want to write it if you don’t want me to :)

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When expressing sets, you may write "inf" for infinity and "U" for union.
GarryVolchara [31]

Answer:

The complement of the given set in interval notation is(-\infty,-5]\cup(6,\infty). It can we written as (-inf,5]U(6,inf).

Step-by-step explanation:

The given set in interval notation is

(−5,6]

It means the set is defined as

A=\{x|x\in R,-5

If B is a set and U is a universal set, then complement of set B contains the elements of universal set but not the elements of set B.

Here, universal set is R, the set set of all real numbers.

U=\{x|x\in R\}

The complement of the given set is

A^c=U-A

A^c=\{x|x\in R,-\infty

Complement of the given set in interval notation is

A^c=(-\infty,-5]\cup(6,\infty)

Therefore the complement of the given set in interval notation is(-\infty,-5]\cup(6,\infty). It can we written as (-inf,5]U(6,inf).

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