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Natasha_Volkova [10]
4 years ago
14

Find the area of this triangle..

Mathematics
1 answer:
DerKrebs [107]4 years ago
6 0

Answer:

<h2>Area =28.8</h2>

Step-by-step explanation:

\frac{bh}{2} \\\frac{8\times 7.2 }{2} \\= 57.5/2\\Area  = 28.8

You might be interested in
Find the value of x in a triangle. x-20, x+10, x-20
tatyana61 [14]

The value of the angles should be. (x-40°). , (x-20°), (½x-10°) , not (x-40°) + (x-20°)+(½x-10°)

Sum of all the interior angles of a triangle is 180°.

So a equation can be made by the given data,

(x-40°) + (x-20°) + (½ x-10°) = 180°

x-40°+x-20°+½x-10° = 180°

2x+½x -60°-10° = 180°

5/2 x - 70° = 180°

5/2 x = 180° + 70°

5/2 x = 250°

x = 250° × 2/5

x = 50° × 2

x = 100°

So the angles are

x-40° = 100°-40° = 60°

x-20° = 100°–20° = 80°

½x-10° = ½(100)° - 10° = 50° -10° = 40°

The answer can be checked by putting the values of the angle we got in the second statement i.e. Sum of all the interior angles of a triangle is 180°.

60° + 80° + 40° = 100° + 80° = 180°

6 0
3 years ago
Another one pls help!!
Mumz [18]

Answer:

21/41

Step-by-step explanation:

Find the total number of students

6+7+2+3+6+8+4+5=41

Then find the students that do not study chemistry

6+7+2+5 = 20

P( not study chemistry) = do not study chem)/total

                                       =21/41

5 0
4 years ago
Please help!<br> 10<br> ∑ 2i-9<br> i=3
pentagon [3]

\sum\limits_{i=3}^{10}(2i-9)=2(3)-9+2(4)-9+2(5)-9+2(6)-9+2(7)-9\\\\+2(8)-9+2(9)-9+2(10)-9\\\\=6-9+8-9+10-9+12-9+14-9+16-9+18-9+20-9\\\\=90-72=18

6 0
3 years ago
AYUDA CON ESTO!!! ALGUIEN PORFAVOR
Gre4nikov [31]

Answer:

Problem 1)  frequency:  160 heartbeats per minute, period= 0.00625 minutes (or 0.375 seconds)

Problem 2) Runner B has the smallest period

Problem 3) The sound propagates faster via a solid than via air, then the sound of the train will arrive faster via the rails.

Step-by-step explanation:

The frequency of the football player is 160 heartbeats per minute.

The period is (using the equation you showed above):

Period = \frac{1}{frequency} = \frac{1}{160} \,minutes= 0.00625\,\,minutes = 0.375\,\,seconds

second problem:

Runner A does 200 loops in 60 minutes so his frequency is:

\frac{200}{60} = \frac{10}{3} \approx  3.33   loops per minute

then the period is: 0.3 minutes (does one loop in 0.3 minutes)

the other runner does 200 loops in 65 minutes, so his frequency is:

\frac{200}{65} = \frac{40}{13} \approx  3.08   loops per minute

then the period is:

\frac{13}{40} =0.325\,\,\,minutes

Therefore runner B has the smaller period

8 0
3 years ago
If anyone could help that would be great / right answer gets brainilest
Keith_Richards [23]
2.69 cm would be the answer
7 0
3 years ago
Read 2 more answers
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