Answer:
centre is 2,-3
radius is 2 units
Step-by-step explanation:
I dont know how to find the equation
above answers are also guesses.
hope u don't mind
Answer:
the answer is a- 60
Step-by-step explanation:
A triangular prism has three rectangular sides and two triangular faces. To find the area of the rectangular sides, use the formula A = lw, where A = area, l = length, and h = height. To find the area of the triangular faces, use the formula A = 1/2bh, where A = area, b = base, and h = height.
Answer:
The area of the circle is 379.94 mm²
Step-by-step explanation:
To solve this problem we need to use the area formula of a circle:
a = area
r = radius = 11 mm
π = 3.14
a = π * r²
we replace with the known values
a = 3.14 * (11 mm)²
a = 3.14 * 121 mm²
a = 379.94 mm²
Round to the nearest tenth
a = 379.94 mm² = 379.9 mm²
The area of the circle is 379.9 mm²
Answer:
x = 29
Step-by-step explanation:
If the equation is :
4.2(9-x)+36=102-2.5(2x+2)
→distribute 4.2 and -2.5 in parenthesis
4.2(9 - x) + 36 = 102 -2.5(2x + 2)
37.8 - 4.2x +36 = 102 -5x -5
→ have terms with x = terms without x
Keep the terms you need the way they are, and move the terms you need on the other side of equal sign with changed sign.
4.2(9 - x) + 36 = 102 -2.5(2x + 2)
37.8 - 4.2x +36 = 102 -5x -5
- 4.2x +5 x = - 37.8 - 36 + 102 - 5
→ Combine like terms
0.8x = 23.2
→ Divide both sides by 0.8
x = 29
Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
