Answer:
1/100 cos100⁰ 100 0.1 tan100⁰
Correct question :
If the perimeters of each shape are equal, which equation can be used to find the value of x? A triangle with base x + 2, height x, and side length x + 4. A rectangle with length of x + 3 and width of one-half x. (x + 4) + x + (x + 2) = one-half x + (x + 3) (x + 2) + x + (x + 4) = 2 (one-half x) + 2 (x + 3) 2 (x) + 2 (x + 2) = 2 (one-half x) + 2 (x + 3) x + (x + 2) + (x + 4) = 2 (x + 3 and one-half)
Answer: (x + 2) + x + (x + 4) = 2 (one-half x) + 2 (x + 3)
Step-by-step explanation:
Given the following :
A triangle with base x + 2, height x, and side length x + 4 - - - -
b = x + 2 ; a = x ; c = x + 4
Perimeter (P) of a triangle :
P = a + b + c
P =( x + 2) + x + (x + 4) - - - (1)
A rectangle with length of x + 3 and width of one-half x
l = x + 3 ; w = 1/2 x
Perimeter of a rectangle (P) = 2(l+w)
P = 2(x+3) + 2(1/2x)
If perimeter of each same are the same ; then;
(1) = (2)
(x + 2) + x + (x + 4) = 2(x+3) + 2(1/2x)
Answer:
b^2-4b+3=0
b²-3x-b+3=0
b(b-3)-1(b-3)=0
(b-3)(b-1)=0
either
b=3 or b=1
.
2n^2 + 7 = -4n + 5
2n²+4n+7-5=0
2n²+4n+2=0
2(n²+2n+1)=0
(n+1)²=0/2
:.n=-1
.
x - 3x^2 = 5+ 2x - x^2
0=5+ 2x - x^2-x +3x^2
0=5+x+2x²
2x²+x+5=0
comparing above equation with ax²+bx +c we get
a=2
b=1
c=5
x={-b±√(b²-4ac)}/2a ={-1±√(1²-4×2×5)}/2×1
={-1±√-39}/2
Answer:
4
Step-by-step explanation:
Yk what all above you can’t go wrong with choosing all of the em tbh cuz u would get one of them righ
