A Pythagorean triple is a set of thre integer numbers, a, b and c that meet the Pythgorean theorem a^2 + b^2 = c^2
Use Euclide's formula for generating Pythagorean triples.
This formula states that given two arbitrary different integers, x and y, both greater than zero, then the following numbers a, b, c form a Pythagorean triple:
a = x^2 - y^2
b = 2xy
c = x^2 + y^2.
From a = x^2 - y^2, you need that x > y, then you can discard options A and D.
Now you have to probe the other options.
Start with option B, x = 4, y = 3
a = x^2 - y^2 = 4^2 - 3^2 = 16 -9 = 7
b = 2xy = 2(4)(3) = 24
c = x^2 9 y^2 = 4^2 + 3^2 = 16 + 9 = 25
Then we could generate the Pythagorean triple (7, 24, 25) with x = 4 and y =3.
If you want, you can check that a^2 + b^2 = c^2; i.e. 7^2 + 24^2 = 25^2
The answer is the option B. x = 4, y = 3
Answer:
y=15
Step-by-step explanation:
The formula for direct variation is
y = kx
5 =k*2
Solve for k
Divide by 2
5/2 = k
y = 5/2 x
Now we substitute x=6
y = 5/2(6)
y = 15
Notice that
(1 + <em>x</em>)(1 + <em>y</em>) = 1 + <em>x</em> + <em>y</em> + <em>x y</em>
So we can add 1 to both sides of both equations, and we use the property above to get
<em>a</em> + <em>b</em> + <em>a b</em> = 76 ==> (1 + <em>a</em>)(1 + <em>b</em>) = 77
and
<em>c</em> + <em>d</em> + <em>c d</em> = 54 ==> (1 + <em>c</em>)(1 + <em>d</em>) = 55
Now, 77 = 7*11 and 55 = 5*11, so we get
<em>a</em> + 1 = 7 ==> <em>a</em> = 6
<em>b</em> + 1 = 11 ==> <em>b</em> = 10
(or the other way around, since the given relations are symmetric)
and
<em>c</em> + 1 = 5 ==> <em>c</em> = 4
<em>d</em> + 1 = 11 ==> <em>d</em> = 10
Now substitute these values into the desired quantity:
(<em>a</em> + <em>b</em> + <em>c</em> + <em>d</em>) <em>a</em> <em>b</em> <em>c</em> <em>d</em> = 72,000
Step-by-step explanation:
a number is a sequence of digits with every position being multiplied by increasing powers of 10.
and the whole number is then the sum of of these little multiplications.
the first position left of the decimal point is for 10⁰ (1). the second position is 10¹ (10). the third position for 10² (100). and so on.
the given number is therefore just the short version of
7×10⁴ + 0×10³ + 8×10² + 1×10¹ + 9×10⁰
and therefore, the digit 1 in this number has the value 1×10¹ = 10, which is also the place value, 10¹ = 10.
