Ask the math master to solve, the algebraic expression of the length of a straight line, 30 oceans a

Let the upper part of the leftmost line be m and the lower part of the line n

n²=d²-c²

m+n)²=b²-a²

e²=m²+(c-a)²

So yes, if you need to continue to help, you can ask for it.

Yes, assuming that the intersection of A and E is A, the intersection of A to C is B, the intersection of C and E is C, so ABC is a right triangle, AC 2=AB 2+BC 2AB=( B 2-A 2)-87 BC=25-A so E=AC= [ B 2-A 2)-87] 2+(25-A) 2 is to use the Pythagorean theorem to construct a right triangle.

Let the intersection point at the top of the arc to the point adjacent to it be x, so (87-x) 2+25 2=87 2 solve 87-x= 6944, and the figure is a right-angled trapezoid, you can get (25-a) 2+( b 2-a 2)- 6944) 2=e 2 to get the expression of e, which represents the root number, and the landlord should take a good look at (b 2-a 2)- 6944 to indicate which section is long.

Is this a graph on the coordinate axis? Is A, C balanced? Do the two lines b and d intersect at the origin of the coordinate axis?

First, use a right triangle to solve the length of the vertical line on the left, and then use a right triangle to solve the length of the vertical line on the right, and the difference between the two is the length of the vertical line sandwiched by AC. Make a vertical line of c at the right end of A, and then use the right triangle obtained at the top right to solve the length of E.

The denominator is rational and chemical, you should have learned it! A= 2-1=1 (2+1)= 2 (2+ 2), B= 2(2-3)= 2 (2+ 3), C= 2(3-2)= 2 (2+3), because 2+ 3>2+ 2> 3+ 2, so C>A>B

Because, so the value of the arithmetic square root of 2 is between.

So the value of a is between, and the value of the arithmetic square root of 2 is between.

Because, so the value of the arithmetic square root of 6 is between.

So the value of c is in between, and the value of b is in between.

So c>a>b.

Suppose that the values of vt and vo are both positive and negative. One positive and one negative. One is zero, two are zero, and the size is divided into these situations, and you can come out

1) A travels 20-18=2 kilometers more per hour than B.

If two people meet at a distance of 2 km from the midpoint, A walks 4 km more than B, so each person walks 4 2 = 2 hours.

Distance = (20 + 18) x 2 = 76 km.

2) 2 hours apart 141 km, and then after 3 hours two cars meet, that is, 3 hours two cars have traveled a total of 141 km, that is, a total of 141 3 = 47 km per hour.

The two cars traveled for a total of 5 hours, and the distance between the two places = 47x5 = 235 kilometers.

t-18t = 2x2 t = 2 hours, s = 20t + 18t = 40 + 36 = 76 km.

2. Let the sum of the speed of A and B be y, and the distance is s.

s-141 = 2y, s = 5y, and the solution is s = 235 km.

(k+2)x=5

x=5/(k+2)

Because the solution is a natural number, it means that k+2 is divisible by 5.

So k+2=5 or k+2=1

The solution yields k = 3 or k = -1

(k 2) x 5, k is -1, 3, -3, -7 (not divisible by 5 when k is another number).

a2+2ab+b2=3,a2-2ab+b2=5

Then the two formulas are added a2+b2=4

Subtract the two formulas by 4ab=-2

So a2+ab+b2=