In △ ABC, AB is equal to AC, BC is equal to 10cm, and the perimeter of the triangle is greater than 34cm and less than 44cm. Find the length range of ab

In △ ABC, AB is equal to AC, BC is equal to 10cm, and the perimeter of the triangle is greater than 34cm and less than 44cm. Find the length range of ab

It is known that ab = AC, triangle ABC is isosceles triangle, because BC = 10, because 44 > perimeter > 34 > AB + AC > 24, and because AB = AC, so 34 > 2Ab > 24, so 17 > AB > 12

(common factor and greatest common factor) 11
Write two numbers with the most common factor of 1
1 \ \ odd number () and even number () 2 \ \ prime number () and composite number ()
3 \ \ composite number () and composite number () 4 \ \ prime number () and prime number ()

1 \ \ odd (1) and even (2) 2 \ \ prime (2) and composite (9)
3-composite (4) and composite (9) 4-prime (2) and prime (3)

If the area ratio of two similar triangles is 4:25, what is the perimeter ratio of the two triangles?
How to calculate?
The waist of an isosceles trapezoid is equal to the upper base and half of the lower base. If the circumference of the isosceles trapezoid is 50, the median line length of the isosceles trapezoid is ()

1. Because the square of the perimeter ratio of similar triangles is equal to the area ratio
So the perimeter ratio is √ 4: √ 25 = 2:5
2. Let the top and bottom be a
Because the bottom is half of the bottom
So bottoms = 2A
Because the waist is equal to the bottom and the trapezoid is isosceles trapezoid
So waist = a
Because 2 waist + upper sole + lower sole = circumference = 50
So 2A + A + 2A = 50
a=10
The bottom is: 2A = 20
The median is (20 + 10) / 2 = 15

The greatest common divisor of two numbers is 6 and the least common multiple is 108. One of them is 12 and the other is 10______ ．

108 △ 12 = 99 × 6 = 54 answer: the other number is 54

The positive half axis of the y-axis at the intersection of the line L1 with the slope k passing through point P (2,3) is at point M. the positive half axis of the x-axis at the intersection of the line L2 with the slope k passing through point P perpendicular to point L1 is at point n
1) Finding the coordinates of points m and n (expressed by K)
2) Find the value range of length C of line segment Mn
3) Finding the value range of ompn area s of quadrilateral

1) Let the equation of line L1: Y-3 = K (X-2)
y=kx-2k+3
Let x = 0
y=3-2k
According to the meaning of the title
3-2k>0
k0
k>-2/3
So - 2 / 3

This is an apple tree
It is helpful for the responder to give an accurate answer

these are apple trees

What quadrant is the vertex of parabola y = x * 2 + 3x

y=x²+3x=(x+ 3/2)² -9/4
Vertex coordinates (- 3 / 2, - 9 / 4)
-3/2

When a natural number greater than 1 is removed from 300243205 and the same remainder is obtained, then the natural number is______ ．

Let this natural number be x, a, B and C be natural numbers, and a be a constant, because when this number is divided by 300243205, we get the same remainder, so ax + a = 300 (1); BX + a = 243 (2); CX + a = 205 (3); (1) - (2) get, (a-b) x = 57, (2) - (3) get, (B-C) x = 38, (1) - (3)

Find the equation of equiaxed hyperbola which passes through point a (3, - 1) and whose symmetry axis is on the coordinate axis

When the focus is on the x-axis, let the standard equation of hyperbola be x2a2 − y2a2 = 1, substitute a (3, - 1) into the equation to get 9A2 − 1A2 = 1, A2 = 8, and the standard equation of hyperbola be x28 − Y28 = 1; (4 points) when the focus is on the y-axis, let the standard equation of hyperbola be y2a2 − x2a2 = 1, and substitute a (3, - 1) into the equation to get 1A2 − 9A2 = 1, A2 = - 8, which does not exist

Two math problems in grade two of junior high school
1. Given that a and B satisfy the equation x = A2 + B2 + 20, y = 4 (2b-a), then the size relation of X and Y is:
A.x=y
C.xy
2. Given x2 + y2-2x + 4y-6z + 14 = 0, find the value of X + y + Z
Note: X2, Y2, A2 and B2 respectively mean the square of a, B, C and D

1. ∵ X-Y = A & sup2; + B & sup2; + 20-8b + 4A = (a + 2) & sup2; + (B-4) & sup2; ≥ 0 ∵ x ≥ y choose B2, ∵ X & sup2; + Y & sup2; - 2x + 4y-6z + 14 = 0 ∵ (x-1) & sup2; + (y + 2) & sup2; - 6Z + 9 = 0. ①: when z = 1.5, x + y + Z = 1-2 + 1.5 = 0.5. ②: when z > 1.5, the value of X + y + Z is uncertain