Answer in simplest radical form would be 7.(:
D 64 if this is right please give me brainlist
We need Pythagoras theorem here
a^2+b^2 = c^2
a, b = legs of a right-triangle
c = length of hypotenuse
Let S=shorter leg, in cm, then longer leg=S+2 cm
use Pythagoras theorem
S^2+(S+2)^2 = (10 cm)^2
expand (S+2)^2
S^2 + S^2+4S+4 = 100 cm^2 [collect terms and isolate]
2S^2+4S = 100-4 = 96 cm^2
simplify and form standard form of quadratic
S^2+2S-48=0
Solve by factoring
(S+8)(S-6) = 0 means (S+8)=0, S=-8
or (S-6)=0, S=6
Reject nengative root, so
Shorter leg = 6 cm
Longer leg = 6+2 cm = 8 cm
Hypotenuse (given) = 10 cm
Answer: The first equation is an equation of a parabola. The second equation is an equation of a line.
Explanation:
The first equation is,

In this equation the degree of y is 1 and the degree of x is 2. The degree of both variables are not same. Since the coefficients of y and higher degree of x is positive, therefore it is a graph of an upward parabola.
The second equation is,

In this equation the degree of x is 1 and the degree of y is 1. The degree of both variables are same. Since both variables have same degree which is 1, therefore it is linear equation and it forms a line.
Therefore, the first equation is an equation of a parabola. The second equation is an equation of a line.
We are given : m∠WYX=(2x−1)° and m∠WYZ=(4x+1)°.
∠WYX and ∠WYZ are complementary.
We know, sum of complementary angles is = 90°.
So, we need to add ∠WYX and ∠WYZ and set it equal to 90°.
m∠WYX + m∠WYZ = 90°.
Plugging values of ∠WYX and ∠WYZ in the above equation, we get
(2x−1)° + (4x+1)° = 90°.
Removing parentheses from both sides,
2x-1 + 4x+1 =90.
Combining like terms,
2x+4x= 6x and -1+1 =0
6x +0 =90.
6x=90.
Dividing both sides by 6.
6x/6 =90/6
x= 15.
Plugging value of x=15.
m∠WYX=(2x−1)° = 2*15 -1 = 30 -1 =29
m∠WYZ=(4x+1)° = 4*15 +1 = 60+1 = 61.
Therefore, ∠WYX=29° and ∠WYZ=61°.