Answer:
6
Step-by-step explanation:
2 + (1/10)(1/5)(1/10)2000 = 2 + (1/500)(2000) = 2 + 4 = 6
Composite numbers are numbers that have more than 2 factors. With that being said, your answer would be 2.
4 is a composite number: 1, 2, 4
6 is a composite number: 1, 2, 3, 6
8 is a composite number: 1, 2, 4, 8
2 is a prime number: 1 and 2
2 is a prime number, not a composite number.
Answer:
x = -1, y = 1
Step-by-step explanation:
6 + 4x - 2y = 0 (1)
-3 - 7y = 10x (2)
From (1)
6 + 4x - 2y = 0 (1)
4x - 2y = -6 (3)
From (2)
-3 - 7y = 10x (2)
10x + 7y = -3 (4)
4x - 2y = -6 (3)
10x + 7y = -3 (4)
Using elimination method
Multiply (3) by 10 and (4) by 4 to eliminate x
40x - 20y = -60
40x + 28y = -12
28y - (-20y) = -12 - (-60)
28y + 20y = -12 + 60
48y = 48
y = 48/48
y = 1
Substitute y = 1 into (3)
4x - 2y = -6 (3)
4x - 2(1) = -6
4x - 2 = -6
4x = -6 + 2
4x = -4
x = -4/4
x = -1
x = -1, y = 1
The value of x in the equation is -1
Answer:
the constant of proportionality k is given by k=y/x where y and x are two quantities that are directly proportional to each other. Once you know the constant of proportionality you can find an equation representing the directly proportional relationship between x and y, namely y=kx, with your specific k.
Step-by-step explanation:
The answer is <span>√x + √y = √c </span>
<span>=> 1/(2√x) + 1/(2√y) dy/dx = 0 </span>
<span>=> dy/dx = - √y/√x </span>
<span>Let (x', y') be any point on the curve </span>
<span>=> equation of the tangent at that point is </span>
<span>y - y' = - (√y'/√x') (x - x') </span>
<span>x-intercept of this tangent is obtained by plugging y = 0 </span>
<span>=> 0 - y' = - (√y'/√x') (x - x') </span>
<span>=> x = √(x'y') + x' </span>
<span>y-intercept of the tangent is obtained by plugging x = 0 </span>
<span>=> y - y' = - (√y'/√x') (0 - x') </span>
<span>=> y = y' + √(x'y') </span>
<span>Sum of the x and y intercepts </span>
<span>= √(x'y') + x' + y' + √(x'y') </span>
<span>= (√x' + √y')^2 </span>
<span>= (√c)^2 (because (x', y') is on the curve => √x' + √y' = √c) </span>
<span>= c. hope this helps :D</span>
