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Sever21 [200]
3 years ago
11

Consider the inequality -5(x+7)<-10. Write and inequality representing the solution for x

Mathematics
1 answer:
uysha [10]3 years ago
5 0
Let's solve your inequality step-by-step.<span><span>−<span>5<span>(<span>x+7</span>)</span></span></span><<span>−10</span></span>Step 1: Simplify both sides of the inequality.<span><span><span>−<span>5x</span></span>−35</span><<span>−10</span></span>Step 2: Add 35 to both sides.<span><span><span><span>−<span>5x</span></span>−35</span>+35</span><<span><span>−10</span>+35</span></span><span><span>−<span>5x</span></span><25</span>Step 3: Divide both sides by -5.<span><span><span>−<span>5x</span></span><span>−5</span></span><<span>25<span>−5</span></span></span><span>x><span>−5

</span></span>Answer:<span>x><span>−5

Good luck mate :P
</span></span><span>
</span>

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Answer:

  3. no; does not have the correct ratio to other sides

  4. (1; right), (2; obtuse), (3; right)

Step-by-step explanation:

3. There are a couple of ways you can go at this.

a) the ratio of sides of a 30°-60°-90° right triangle (half an equilateral triangle) is 1 : √3 : 2. Here, the ratio of sides is ...

  (√30/2) : √15 : √30 = 1 : √2 : 2 . . . . MQ is <em>not</em> the height

b) the height is perpendicular to the base, so if MQ were the height, the sides of the triangle would satisfy the Pythagorean theorem:

  LQ² +MQ² = LM²

  (√30/2)² +(√15)² = (√30)²

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The length MQ cannot be the height of ∆LMN. (It is too short.)

__

4. To see if these lengths form a right triangle, you can test them in the Pythagorean theorem relation. In the attached we have reformulated the equation so it gives a value of 0 if the triangle is a right triangle:

  c² = a² +b² . . . . . Pythagorean theorem

  c² -a² -b² = 0 . . . . rewritten

If the value of c² -a² -b² is not zero, then the side lengths do not form a right triangle. The attached shows this math performed by a graphing calculator. The results are ...

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<em>Additional comment</em>

The attached table demonstrates an artifact of computer arithmetic. Not all numbers can be represented exactly in a computer, so a difference that is expected to be zero using exact arithmetic may be slightly different from zero when computed by a computer or calculator. Here, we see a difference of about 6×10^-14 when zero is expected. On most calculators (with 10, or 12 displayed digits), this would be displayed as 0.

The same calculation done "by hand" gives ...

  (√425)² -5² -20² = 425 -25 -400 = 0 . . . Triangle 1 is a <em>right triangle</em>

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