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Lapatulllka [165]
3 years ago
6

Without actually drawing the figure, could you form a triangle using side lengths 7, 8, and 18 units? Why or why not.

Mathematics
2 answers:
Marysya12 [62]3 years ago
7 0

Answer:

Step-by-step explanation:

ZanzabumX [31]3 years ago
6 0
Here is an example you can learn from.

#1:   Is it possible to form a triangle with the given side lengths? If not, explain why not. 1. 5 cm, 7 cm, 10 cm SOLUTION:   The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Yes; 5 + 7 > 10, 5 + 10 > 7, and 7 + 10 > 5 ANSWER:   Yes; 5 + 7 > 10, 5 + 10 > 7, and 7 + 10 > 5

2. 3 in., 4 in., 8 in. SOLUTION:   No; . The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ANSWER: 3+4*8

3. 6 m, 14 m, 10 m SOLUTION:   Yes; 6 + 14 > 10, 6 + 10 > 14, and 10 + 14 > 6 .  The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ANSWER:   Yes; 6 + 14 > 10, 6 + 10 > 14, and 10 + 14 > 6 

4. MULTIPLE CHOICE If the measures of two sides of a triangle are 5 yards and 9 yards, what is the least possible measure of the third side if the measure is an integer? A 4 yd B 5 yd C 6 yd D 14 yd SOLUTION:   Let x represents the length of the third side. Next, set up and solve each of the three triangle inequalities. 5 + 9 > x, 5 + x > 9, and 9 + x > 5 That is, 14 > x, x > 4, and x > –4. Notice that x > –4 is always true for any whole number measure for x. Combining the two remaining inequalities, the range of values that fit both inequalities is x > 4 and x < 14, which can be written as 4 < x < 14. So, the least possible measure of the third side could be 5 yd.  The correct option is B. ANSWER:   B 
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What's the answer to this please help. :)
Alexxandr [17]
I think the answer is 32 if I did my math right sorry if it wrong
6 0
3 years ago
Read 2 more answers
Working independently, Tina can do a certain job in 12 hours. Working independently, Ann can do the same job in 9 hours. If Tina
Leona [35]

Answer:

Ann would need 3 h to complete the remainder of the Job.

Step-by-step explanation:

Step 1.

The "speed" that each girl has to do the work is determined as:

  • Tina: \frac{1}{12h}, it means that Tina complete one twelveth of the job each hour.
  • Ann: \frac{1}{8h}, it means the Ann completes one eighth of the job each hour.

Step 2.

The amount of work done by Tina in 9 hours, is obtain multiplying by the "speed" calculated in step 1:

Amount of work done= 8 h*\frac{1}{12h} = \frac{2}{3}. It means that <em>Tina completes two thirds of the work in 9 h.</em>

Step 3.

The remaining job is calculated as 1-\frac{2}{3}=\frac{1}{3}. It means that <em>still remains is one third of the job to be completed</em>.

Step 4.

The time required for Ann to complete the job is calculated dividing the remaining of the job by the "speed" of Ann to do the job.

\frac{\frac{1}{3}}{\frac{1}{9h}}=3h. It means that <em>Ann would complete the job in another 3 hours</em>.

3 0
3 years ago
43 base five to base seven
solong [7]
First convert to decimal:

43_5=4\cdot5^1+3\cdot5^0=20+3=23

Now convert this to base 7:

\dfrac{23}7=3+\dfrac27
\dfrac27=0+\dfrac27
\implies23=43_5=32_7
7 0
3 years ago
An arc subtends a central angle measuring \dfrac{\pi}{2}
Agata [3.3K]

Arc Length is 1/4th of the circumference .

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

Here we need to find fraction of the circumference is this arc when An arc subtends a central angle measuring \dfrac{\pi}{2} radians ! Let's find out :

We know that circumference of an arc subtending a central angle of x is :

⇒ Arc = \frac{Angle}{360}(2\pi r)

⇒ Arc = \dfrac{\frac{\pi}{2}}{360}(2\pi r)

⇒ Arc = \frac{90}{360}(2\pi r)

⇒ Arc = \frac{90}{90(4)}(2\pi r)

⇒ Arc = \frac{1}{4}(2\pi r)

⇒ Arc = \frac{1}{4}(Circumference)

Therefore , Arc Length is 1/4th of the circumference .

4 0
3 years ago
What is 46 2/3% of 28?
lilavasa [31]

46 2/3 %

= (46 + 2/3)/100

= (138/3 + 2/3)/100

= (140/3)/100

= 140/300

= 7/15

So it's

7/15 * 28

= 196/15

= 15 1/13

6 0
3 years ago
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