Answer:
16
Step-by-step explanation:
8x2
Answer:
80
Step-by-step explanation:
Answer:
Step-by-step explanation:
a.
2y - 3x = 5
2(-2) - 3x = 5
-4 - 3x = 5
-3x = 9
x = -3
(-3,-2) is another solution
b.
(-1,1)
2y - 3x = 5
2(1) - 3(-1) = 5
2 + 3 = 5
5 = 5
true, the point is a solution to the equation because the equation is true after substituting x and y with the point (-1,1)
(4,1)
2y - 3x = 5
2(1) - 3(4) = 5
2 - 12 = 5
-10 = 5
not true, the point is not a solution to the equation because the equation is not true after substituting x and y with the point (4,1)
c. You can use the points from the given (-1,1) and (-3,-2) to form a line. You then shade whichever half solves the solution using points on the graph.
You only have to apply the theorem of Pythagoras here. Remember the square on the hypotenuse (the longest side) is equal to the sum of the squares on the other two sides :
1. AB is the hypotenuse, so, according to the theorem we can write :
AB² = AC² + CB²
c² = 5² + 4²
c²= 25 + 16
c² = 41
applying the square root of 41 we get :
c ≈ 6.40 rounded to the hundred
The next cases are exactly the same thing so there is no need for explanation :
2.
AB is the hypotenuse here because it is the biggest side clearl :
AB² = AC² + CB²
25² = 15² + b²
Thus
b² = 25² - 15²
we just subtracted 15² on each side of the equation
b² = 625 - 225
b² = 400
applying the square root of 400 we get
b = √400 = 20
So AC = 20
3. The longest side is clearly AB = 60
So
AB² = AC² + CB²
60² = 40² + a²
subtracting 40² on each side of the equation we get :
a² = 60² - 40²
I let you finish this using your calculator and doing exactly like the previous cases
4.
AB is the hypotenuse,
AB² = AC² + CB²
23² = b² + 14²
Subtracting 14² from each side of the equation we get
b² = 23² - 14²
5.
AB is the biggest side :
AB² = AC² + CB²
29² = 23² + a²
We subtract 23² on each sides of the equation :
a² = 29² - 23²
You can finish with your calculator
6.
AB² = AC² + BC²
78² = b² + 30²
subtraction...
b² = 78² - 30²
Good luck :)
Answer:

Step-by-step explanation:

= 91
*
= 