A ladder is 25m long, leaning against a wall. The low end of the ladder is 7m away from the wall. How high is the top of the ladder from the ground?

A ladder is 25m long, leaning against a wall. The low end of the ladder is 7m away from the wall. How high is the top of the ladder from the ground?

Distance from top of ladder to ground = under root sign (25 ^ 2-7 ^ 2) = 24m or
25²-7²=24²=24

A ladder is 25 meters long, leaning against a wall. The bottom of the ladder is 7 meters away from the wall. (1) how high is the top of the ladder from the ground? (2) If the top of the ladder slides 4 meters to a ', how many meters does the bottom of the ladder slide horizontally?

(1) A: the top of the ladder is 24 meters away from the ground. (b) if BA ′ = 20 meters, BC ′ = 252 − 202 = 15 meters, then CC ′ = 15-7 = 8 meters. A: the bottom of the ladder slides 8 meters horizontally

4. A ladder is 25 meters long, as shown in the picture, leaning against a wall, and the bottom of the ladder is 7 meters away from the wall
(1) How high is the top of this ladder from the ground?
(2) If the top of the ladder falls 4 meters, how many meters does the bottom of the ladder slide horizontally?

(1) Height ^ 2 = 25 * 25-7 * 7 = 576
Height = root 576 = 24m
The top of the ladder is 24 meters above the ground
(2) At this time, the height is 20 meters,
Distance from the lower end to the wall = root (25 * 25-20 * 20) = 15m
Square distance passed = 15-7 = 8M
The bottom of the ladder slid 8 meters horizontally

Factorization. Ax & # 178; + 2axy + ay & # 178; fast

=a（x+y）^2

Factorization: M & # 178; - 4N & # 178; + 2m-4n X & # 178; - 2XY + Y & # 178; + ax ay

m²-4n²+2m-4n
=(m+2n)(m-2n)+2(m-2n)
=(m+2n+2)(m-2n)
x²-2xy+y²+ax-ay
=(x-y)²+a(x-y)
=(x-y+a)(x-y)

Factorization of (AX + by) & #178; + (ay BX) & #178; + 2 (AX + by) (ay BX)

Hello, this question is actually very simple
Let m = (AX + by), n = (ay BX), then the original formula = M & # 178; + n & # 178; + 2Mn, that is, (AX + by + ay BX) &# 178;
Hope to help you... Please adopt

(AX + by) & #178; + (ay BX) & #178; + 2 (AX + by) (ay BX) a (AB + BC + AC) - ABC use the factorization method

（ax+by）²+（ay-bx）²+2（ax+by）（ay-bx）
＝（ax+by+ay-bx）²
a（ab+bc+ac）-abc
=a(ab+bc+ac-bc)
=a(ab+ac)
=a²(b+c)

Solve the equation x & # 178; - (3m-1) + 2m & # 178; - M = 0 with respect to X

x²-(3m-1)+2m²-m=0
x²-(3m-1)+m（2m-1）=0
（x-m）（x-2m+1）=0
X = m or x = 2m-1

It is known that the univariate quadratic equation MX & # 178; - (2m-1) x + (m-2) = 0 (M > 0)
If the two real roots of this equation are respectively X &;, X &;, and (X &; - 3) (X &; - 3) = 5m, find the value of M

x1+x2=(2m-1)/m
x1x2=(m-2)/m
x1x2-3(x1+x2)+9=5m
(m-2)/m-3(2m-1)/m=5m-9
(m-2-6m+3)/m=5m-9
1-5m=5m²-9m
5m²-4m-1=0
1 -1
5 1
(m-1)(5m+1)=0
m=1 m=-1/5
b²-4ac
=(2m-1)²-4m(m-2)
=4m²-4m+1-4m²+8m
=4m+1>0
m>-1/4
∴m=1 m=-1/5

If α and β are two of the equations X & # 178; + (2m-1) x + 4 = 0, and α < 2 < β, the value range of M is obtained

X & # 178; + (2m-1) x + 4 opening upward
Two roots on both sides of x = 2
We can see by drawing
x=2,x²+（2m-1）x+4