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ASHA 777 [7]
3 years ago
8

Solve the problems.

Mathematics
1 answer:
RoseWind [281]3 years ago
6 0

Answer:

????

Step-by-step explanation:

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Given the lengths of two sides of a triangle, find the range for the length of the third side. Write an inequality.
djverab [1.8K]

Answer:

| a - b | < length of third side < a + b

Step-by-step explanation:

Visualize the two given sides of the triangle (let's call then a and b), joined at the vertex of the triangle, and forming an angle. We can join the other free end of these two segments, with another segment whose length would vary according to how tiny or large the angle is. We can spread the aperture of the angle they form as much as we can just below  (not reaching this angle measure, because in such case, there will be no triangle of tangible area. In such case, the length of the joining segment will be limited by the addition of the two sides:

length of third side < a + b

In the case the aperture of the angle formed by the two given sides is diminished as much as possible to still form a measurable triangle, the angle has to be just larger than zero, and in such case, the segment joining the other to ends of a and b would be just larger than the absolute value of the difference between a and b:

length third side > | a - b|

These are the two extreme cases, and the length of the third side must be within these limits.

5 0
3 years ago
Read 2 more answers
Fill in the blank to make the expression a perfect square.<br> v^2 - 18v + ?
Oliga [24]

Answer: (v - 9)^2 - 81

Step-by-step explanation: Use the formula (b/2)^2 in order to create a new term to complete the square.

Hope this helps! :) ~Zane

5 0
3 years ago
The sum of a number and 4 is 6.5
ioda

Answer:

2.5

Step-by-step explanation:\

We can solve it like this

4 + x = 6.5

6.5 - 4 = x

6.5 - 4 = 2.5

SO

4 + 2.5 = 6.5

3 0
3 years ago
Read 2 more answers
On a multiple choice test, if you randomly guessed on three questions, then what is the probability you got at least one of them
snow_lady [41]

Answer:

Number of Questions =3

Probability of giving a correct answer

                              =\frac{1}{3}

Probability of giving two correct answers

                       =\frac{2}{3}

Probability of giving all correct answers

            =\frac{3}{3}\\\\=1

Probability that at least one of them is correct

         =_{1}^{3}\textrm{C}\\\\=\frac{3!}{(3-1)! \times1!}\\\\=3 \text{ways}\\\\=\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \\\\=\frac{1}{27}

Probability that two of them is correct

         =_{2}^{3}\textrm{C}\\\\=\frac{3!}{(3-2)! \times2!}\\\\=3 \text{ways}\\\\=\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \\\\=\frac{8}{27}

Probability that all of them is correct

         =_{3}^{3}\textrm{C}\\\\=\frac{3!}{(3-3)! \times3!}\\\\=1 \text{way}\\\\=1

So, Required probability

         =\frac{1}{27} \times \frac{8}{27} \times 1\\\\=\frac{8}{729}

6 0
3 years ago
Can you solve the system of equations by elimination<br> 2x + 3y = 16<br> 5x - 2y = 21
pogonyaev
The values of x and y are 5 and 2 respectively

5 0
3 years ago
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