Seventh grade math questions are in a hurry, in a hurry

c. Obtuse triangle.

Angle a = 2 angle b = 3 angle c

Angle A = 3 Angle C, Angle B 3 2 Angle C

Angle A angle B angle C 3 angle C 3 2 angle C angle C 180 degrees.

Angle C 360 11 degrees.

Angle A 1080 11 degrees 90 degrees.

So the triangle abc is an obtuse triangle.

Let the angle of angle a be x, then the angle b is 2x, and the angle c is 3x, and the sum of the inner angles of the triangle is 180 degrees.

So. x+2x+3x=180

Solution. x=30

The triangle is a right-angled triangle.

You can set A to 6x, B to 3x, and C to 2x.

The column equation is: 6x+3x+2x=180

Solution: x=180 11

So a = 1080 11, greater than 90 degrees, so this triangle is an obtuse triangle.

Set the angle a=x, then the angle b=2x, and the angle c=3x

x+2x+3x=180

So x=30

then 2x=60

3x=90 so b, right triangle.

Let the angle c be x, then the angle b is 2x, and the angle a is 3x;

Yes, x+2x+3x=180°

x=30° so the angles are 30, 60, 90, so the triangle is right-angled.

Everyone has figured it out, but I don't understand, the obtuse triangle is right. You can substitute the answer to the right triangle, which does not match the angle a = 2 angle b = 3 angle c

Key: Puppy running time = A and B meeting time.

Speed: 6m s

The dog ran: 6*80=480m

The above is definitely true, but I found the following question, can you see if the question is missing?

Two students, A and B, started from a 400-meter circular [a certain point on the track with their backs, jogging at a speed of 2 meters and 3 meters per second respectively, 6 seconds later, a puppy ran from A at a speed of 6 meters per second to B, and after meeting B, it has run to A at a speed of 6 meters per second.

74*6=444m.

It takes 400 (2+3)=80 seconds for A and B to meet, and the puppy runs a total of 80-6=74 seconds, so the puppy runs a total of 74*6=444 meters.

Uh, it's so simple... The triangle sab and the triangle dcb are congruent triangles... sa=cd

The triangle cdb is congruent with the triangle abs, and cd is equal to as

It is expressed as: (1) y=15-5x a solution: x=1, y=10

2) y=(3 4)x-3 a solution: x=0, y=-3; It is also possible to write a lot, for example: x=4, y=0

The main use is factorization in 7th grade mathematics.

Choose c, you look at the denominator of the fractional equation, (x-2) (x-4) = x squared - 6x+8, multiply both sides of the equation by (x-2) (x-4) at the same time to get a one-dimensional equation about x, and solve x is equal to 6. It is a multiple-choice question that you really can't bring in and calculate it all over again, what can be satisfied is the correct answer, and if it is a big question, you have to check it after solving it to prevent the denominator from being zero.

Choose C 4th question, the denominator is not equal to 0, exclude AB, and then substitute C into the question, just right.

Or factor the first denominator and split it into the product of the second and third denominators.

For multiple-choice questions, don't go through the process, just use the elimination method.

。。。Isn't that the second volume of the second year of junior high school?。。。 Is it the first year of junior high school?? Now children learn so fast...

Set the side length of the window is x meters, then.

The area of the semicircle = *(x 2) 2*1 2

From the question: x 2+ * x 2) 2*1 2=2

x²+πx²/8=2

x²(1+π/8)=2

x =2 (1+ 8)=16 (8+ )x= [16 (8+)And because the distance must not be negative, the width of the window is about meters.