It's a simultaneous equation:
Steps:
1.Number the equations..
a+b=77 -1
a-b=13 -2
2. Choose what variable you want to use. In this case I would use the "b". Since the signs in front of the "b's" are different, add the two equations together
a + b = 77
+ + +
a (-b) = 13
Which gives;
2a = 90
Then solve to find a:
2a=90
a= 90/2
a=45
3.Then plug the "a" value into any of the original equations to find the "b" value. I would use equation 1 since the all the variables are positive.
a + b = 77
(45) + b = 77
b=77-45
b=32
4.Solution
a=45
b=32
Answer: 35 inches.
Step-by-step explanation:
We know that:
hypotenuse = 5*y in
cathetus 1 = (x + 8) in
cathetus 2 = (x + 3) in
The perimeter of the triangle is 76 inches, then:
5*y + (x + 8) + (x + 3) = 76
5*y + 2*x + 13 = 76
We also know that the length of the hypotenuse minus the length of the shorter leg is 17 in.
The shorter leg is x + 3, then:
5*y - (x + 3) = 17
Then we have the equations:
5*y + 2*x + 11 = 76
5*y - (x + 3) = 17
With only these two we can solve the system, first we need to isolate one of the variables in one of the equations, i will isolate x in the second equation.
x = 5*y - 3 - 17 = 5*y - 20
x = 5*y - 20
Now we can replace this in the other equation, we get:
5*y + 2*x + 11 = 76
5*y + 2*(5*y - 20) + 11 = 76
15*y - 40 + 13 = 76
15*y - 29 = 76
15*y = 76 + 29 = 105
and remember that the hypotenuse is equal to 5*y, then we want to get:
3*(5*y) = 105
5*y = 105/3 = 35
5*y = 35
Then te length of the hypotenuse is 35 inches.
Okkk so u need to use the Pythagorean Theorum.
a=x
b=1
c=5
so...
Solve that
Answer:
ok so the earth radius is 1534mi so if that is the radius than the area is 7.39×10^6so then you divide that into 3 pieces so each piece would be 2,463,333.3333 the nearest million would be 2,000,000
Step-by-step explanation:
Answer:
0.037
Step-by-step explanation:
If the hitter makes an out 72% of the time, then the probabilty that he makes hit in 1 at-bat is 
and the probabilty that he doesn't make hit in 1 at-bat is 
The probability that the hitter makes 10 outs in 10 consecutive at-bats, assuming at-bats are independent events is
