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Rina8888 [55]
3 years ago
13

Need help please hhhhhhhhhhhhhhhhhhhheeeeeeeeeeeeeeeeeeeeeeeelllllllllllllllllllllllllpppppppppppp

Mathematics
2 answers:
never [62]3 years ago
4 0

An isosceles triangle has two identical angles.

The three angles need to add up to 180 degrees.

If one angle is 100 then the other two angles added together need to equal 180-100 =‘80 degrees.

Since two angles need to be the same the other angles are 80 / 2 = 40 degrees each.

The answer is 40

Fittoniya [83]3 years ago
3 0

Answer:

A. first answer on the left.

Step-by-step explanation:

If one of the angles is 100 then if the triangle is isosceles then the two remaining angles have to be either 100 and - 20 (which is impossible because at this level of math there is no such thing as an angle with a measure of - 20).

The minus 20 comes from every triangle having 180 degrees. 100 + 100 adds to 200, which means the third angle has to be the impossible measure of - 20.

The other possibility is that the given angle of 100 is the top angle and that leaves 80 degrees for the two isosceles angles.

2x = 80

x = 80/2

x = 40

The only possible answer is 40 - 40 - 100

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PLEASE HELP WILL MARK BRAINLIEST!!(plz show work too)
trapecia [35]

\huge \bf༆ Answer ༄

Let the capacity of bus be x students

And van be y students, now ;

From the given statements we get two equations ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260 \:   \:  \:  \: \:  (1)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:13x + 5y = 670 \:  \:  \:  \:  \: (2)

multiply the equation (2) with 2 [ it won't change the values ]

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260 \: \:  \:  \:   \:  \:  \:  \:  \: (1)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:26x + 10y = 1340 \:  \:  \:  \:  \: (3)

Now, deduct equation (1) from equation (3)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:26x + 10y - 2x - 10y = 1340 - 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:24x = 1080

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:x = 1080 \div 24

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:x = 45

Therefore each bus can carry (x) = 45 students

Now, plug the value of x in equation (1) to find y ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:(2 \times 45) + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:90 + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:10y = 260 - 90

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:10y = 170

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y = 170 \div 10

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y = 17

Hence, each van can carry (y) = 17 students in total.

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