Answer:
12<x<16
Step-by-step explanation:
if c^2 > a^2 + b^2
then the triangle is obtuse
since 20 is the longest side
20^2 > x^2 + (x+4)^2
simplify
400 > x^2 + (x+4)(x+4)
FOIL
400 > (x^2 ) + (x^2 + 4x+4x+16)
combine like terms
400 > (2x^2 + 8x+16)
divide by 2
400/2 > 2x^2 /2 + 8x/2 + 16/2
200 > x^2 + 4x +8
subtract 200 from each side
0> x^2 + 4x +8-200
0> x^2 +4x-192
Factor
0 > (x-12) ( x+16)
using the zero product property
0> x-12 0 > x+16
12>x -16>x
x must be greater than 12
we know the longest side is 20
x+4 < 20
subtract 4
x< 16
x>12 and x < 16
12<x<16
4) 4 1/5
5) 6 2/9
6) -6
To solve for these, just divide the most you can and then put the amount that you couldn’t divide anymore on top of the fraction. For example, 4 is the farthest I could divide and I put the 1 as a fraction of 5 as that is what was left.
S is the set of prime numbers that are less than 15. The correct solution in roster form is {1,3,5,7,9,11,13}. The answer is letter C. The rest of the choices do not answer the question above.
Answer:
X= - 44
18x+3x+4-5=5(4x9
Collect like terms
21x+4-5=5(4x-9
More variable to the left hand side and change its sign
21x-20x-1=-45
Correct like terms
X=-45+1
Which is x=-44
Answer:
-a+-2b+4c
Step-by-step explanation:
Combine like terms
-4a+ 3a= -a
b+ -3b= -2b
3c+ c= 4c
