The answer for this problem is 1.7 I think I’m not sure
Answer:
A, C, D
Step-by-step explanation:
Consider triangles NKL and NML. These triangles are right triangles, because
In these right triangles:
- reflexive property;
- given
Thus, triangles NKL and NML by HA postulate. Congruent triangles have congruent corresponding parts, so
![\overline{KN}\cong \overline{NM}\\ \\\overline{KL}\cong \overline{LM}\ [\text{option D is true}]](https://tex.z-dn.net/?f=%5Coverline%7BKN%7D%5Ccong%20%5Coverline%7BNM%7D%5C%5C%20%5C%5C%5Coverline%7BKL%7D%5Ccong%20%5Coverline%7BLM%7D%5C%20%5B%5Ctext%7Boption%20D%20is%20true%7D%5D)
Since
then
![7x-4=5x+12\\ \\7x-5x=12+4\\ \\2x=16\\ \\x=8\ [\text{option A is true}]\\ \\MN=KN=7\cdot 8-4=56-4=52\ [\text{option C is true}]](https://tex.z-dn.net/?f=7x-4%3D5x%2B12%5C%5C%20%5C%5C7x-5x%3D12%2B4%5C%5C%20%5C%5C2x%3D16%5C%5C%20%5C%5Cx%3D8%5C%20%5B%5Ctext%7Boption%20A%20is%20true%7D%5D%5C%5C%20%5C%5CMN%3DKN%3D7%5Ccdot%208-4%3D56-4%3D52%5C%20%5B%5Ctext%7Boption%20C%20is%20true%7D%5D)
Option B is false, because KN=52 units.
Option E is false, because LN is congruent KN, not LM
for 6
in a
given
slope(m)=1
point=(2,-4)=(x1,y1)
we know
equation of straight line
y-y1=m(x-x1)
y+4=1(x-2)
y+4=x-2
therefore x-y-6=0 is the required equation
substituting with ax+by+c=0 we get
a=1
b= -1
c= -6
Answer:
2.8a²+0.9a - 1.2
Step-by-step explanation:
Given the expression 0.3(3a-4)-0.05(8a)(-7a)
Expand using the distributive law
0.3(3a-4)-0.05(8a)(-7a)
0.3(3a)-0.3(4)-0.05(8a)(-7a)
0.9a-1.2 - 0.05(8a)(-7a)
0.9a - 1.2 + 2.8a²
Rearrange
2.8a²+0.9a - 1.2
Hence the required expression is 2.8a²+0.9a - 1.2
