Basic Algebra Problems And Answers
Algebra 2: Basic Word Problems and Solving for an Indicated Variable-10 Pts.?
Hi,
I need help figuring out the answers to these problems:
One side of a triangle is 1 in. longer than the shortest side and is 1 in. shorter than the longest side. The perimeter is 17 in. *Find the dimensions of the triangle* (so whatever the equation is, I’m assuming the end should say “=17″
The sides of a triangle are in the ratio 3 : 4 : 5. What is the length of each side if the perimeter of the triangle is 30 cm?
Last one…
Solve for h: v = s^2 + 1/2sh
I would appreciate any help you can give me in a timely manner. I respect hard work and will gladly take a few seconds out of my day to grant you 10 points and express my gratitude to you.
Thanks again.
Thanks for the help, I’ll give you best answer within 24 hours.
x = shortest side and y = longest side. Let z be the other side.
From information given, z = x + 1 and z = y – 1
giving x = z – 1 and y = z + 1
Perimeter ————–> x + y + z = 17
(z – 1) + (z + 1) + z = 17
3z = 17
z = 17/3
z = 5 2/3
So x = z – 1 = 5 2/3 – 1 = 4 2/3
and y = z + 1 = 5 2/3 + 1 = 6 2/3
Sides are 4 2/3 in, 5 2/3 in and 6 2/3 in
[ 4 2/3 + 5 2/3 + 6 2/3 = 15 + 6/3 = 17]
Sides are in ratio 3 : 4 : 5
Number of parts = 3 + 4 + 5 = 12
Shortest side = 3 X (12 / 30) = 3 X (5 / 2) = 7.5
Next side = 4 X (12/ 30) = 4 X (5/2) = 10
Longest side = 5 X (12 / 30) = 5 X (5 /2) = 12.5
Sides 7.5 cm, 10 cm and 12.5 cm.
v = s^2 + 1/2 sh
Subtract s^2 from each side
v – s^2 = sh / 2
Multiply each side by 2
2(v – s^2) = sh
Divide each side by s
h = 2(v – s^2) / s
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