I saw the image that should have been posted with this word problem.
The rolls of wrapping paper is cylindrical in shape.
Roll A : length = 30 inches ; diameter = 3 inches
Roll B : length = 24 inches ; diameter = unknown
Lateral surface area = 2 π r h
Lateral Surface area of Roll A = Lateral Surface area of Roll B
2 * 3.14 * 1.5in * 30in = 2 * 3.14 * r * 24 in
282.60 in² = 150.72 in * r
282.60 in² / 150.72 in = r
1.875 in = r
diameter is 2 times the radius = 1.875 in * 2 = 3.75 inches
The diameter of Roll B is 3.75 inches.
Solve for b.
b^2= x+4ac
So b=sq root of x+4ac
Answer:
The completed proof is presented as follows;
The two column proof is presented as follows;
Statements 
Reason
1. 
║ 
, J is the midpoint of 
1. Given
2. ∠IHJ ≅ ∠JLK 2. Alternate angles are congruent
3. ∠IJH ≅ ∠KJL 3. Vertically opposite angles
4. 
≅ 
4. Definition of midpoint
5. ΔHIJ ≅ ΔLKJ 5. By ASA rule of congruency
Step-by-step explanation:
Alternate angles formed by the crossing of the two parallel lines and , by the transversal are equal
Vertically opposite angles formed by the crossing of two straight lines 
and are always equal
A midpoint divides a line into two equal halves
Angle-Side-Angle, ASA rule of congruency states that two triangles ΔHIJ and ΔLKJ, that have two congruent angles, ∠IHJ in ΔHIJ ≅ ∠JLK in ΔLKJ and ∠IJH in ΔHIJ ≅ ∠KJL in ΔLKJ, and that the included sides between the two congruent angles is also congruent ≅ , then the two triangles are congruent, ΔHIJ ≅ ΔLKJ.
Answer:
no solution
Step-by-step explanation:
49^3x = 343^2x+1
49 = 7^2 and 343 = 7^3
7^2 ^3x = 7^3 ^2x+1
we know that a^b^c = a^ (b*c)
7^(2 *3x) = 7^(3 *(2x+1))
7^(6x) = 7^(6x+3)
The bases are the same so the exponents have to be the same
6x = 6x+3
Subtract 6x from each side
0 =3
This is never true so there is no solution
Answer:
Step-by-step explanation:
Remember PEMDAS
multiply 4 and 8
32+3
add 32 and 3
35
