Since x=x, this is an isosceles right triangle. By the Pythagorean Theorem:
h^2=a^2+b^2 (the hypotenuse squared is equal to the sum of the squared sides)
5^2=x^2+x^2
25=2x^2
2x^2=25
x^2=25/2
x=√(25/2)
x=5/√2 now if we rationalize the denominator...
x=(5√2)/(√2√2)
x=(5√2)/2
58 is 5 tens (5*10) and 8 ones (8*1)
Answer with explanation:
The equation of line is, y= -x +3
→x+y-3=0---------(1)
⇒Equation of line Parallel to Line , ax +by +c=0 is given by, ax + by +K=0.
Equation of Line Parallel to Line 1 is
x+y+k=0
The Line passes through , (-5,6).
→ -5+6+k=0
→ k+1=0
→k= -1
So, equation of Line Parallel to line 1 is
x+y-1=0
⇒Equation of line Perpendicular to Line , ax +by +c=0 is given by, bx - a y +K=0.
Equation of Line Perpendicular to Line 1 is
x-y+k=0
The Line passes through , (-5,6).
→ -5-6+k=0
→ k-11=0
→k= 11
So, equation of Line Parallel to line 1 is
x-y+11=0
The function is (-x+3)/ (3x-2) and we get f(1)=1 and differentiation is f'(x)=-7/ (9x²- 12x+4).
Given that,
The function is (-x+3)/ (3x-2)
We have to find f(1) and f'(x).
Take the function expression
f(x)= (-x+3)/ (3x-2)
Taking x as 1 value
f(1)= (-1+3)/(3(1)-2)
f(1)=2/1
f(1)=1
Now, to get f'(x)
With regard to x, we must differentiate.
f(x) is in u/v
We know
u/v=(vu'-uv')/ v² (formula)
f'(x)= ((3x-2)(-1)- (-x+3)(3))/ (3x-2)²
f'(x)= ((-3x+2)-(-3x+9))/ 9x²- 12x+4
f'(x)=(-3x+2+3x-9)/ 9x²- 12x+4
f'(x)=2-9/ (9x²- 12x+4)
f'(x)=-7/ (9x²- 12x+4)
Therefore, The function is (-x+3)/ (3x-2) and we get f(1)=1 and differentiation is f'(x)=-7/ (9x²- 12x+4).
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