Answer: the awnser is 9, 40, 41 mark me brainlist
Step-by-step explanation:
Sum of the squares of the two side = Square of longest side
a² + b² = c²
a)
So let's check 7, 24, 25
Is 7² + 24² = 25² ?
7*7 + 24*24
49 + 576
=625.
Let us perform the other side 25²
25² = 25 * 25 = 625
Therefore the left hand side = Right hand side.
Therefore 7, 24, 25 is a Pythagorean Triple
b)
Let's check 9, 40, 41
Is 9² + 40² = 41² ?
9² + 40²
9*9+ 40*40
81 + 1600
=1681
Let us perform the other side 41²
41² = 41 * 41 = 1681
Therefore the left hand side = Right hand side.
Therefore 9, 40, 41 is a Pythagorean Triple.
Read more on Brainly.com - brainly.com/question/1471164#readmore
Answer:
3/2 and 3/4 are both not multiples of 1/3.
Step-by-step explanation:
They have to have 3 as the denominator (or the number below the line) to be a multiple of 1/3. For example: 1/3, 2/3, 3/3.
Hope you understand that! :)
This one can besolved using system of equations. From the first statement we can assume that
length = 2 * width. The area of rectange equals product of it's width and length. By knowing that and following to the second statement we get the second equation:
(3 * width) * (length - 5) = width * length + 4. Let's mark the
width as
x and the
length as
y and write down our equations system:
Using substituion method, lets replace every
y with
2x within second equation and solve it:
Now lets find the discriminant in order to solve this quadratic equation:
The second root is negative, so we ignore it as
x represents width which can't be negative.
Now, using found root, let's find
y value from the first equation:
So,
the width is equal 4 (and the length is equal 8).
To solve for any given number of variables, you need that number of equations to solve for all variables. When there are only two variables you only need to equations to solve for them both. You eliminate one variable at a time, order does not matter.
y-x=6 and y+x=-10
Since one equation has -x and the other +x, you can simply add the two equations together to eliminate x...
(y-x=6)+(y+x=-10)
y+y-x+x=6-10
2y=-4
y=-2, now you can use this value of y in either original equation to solve for x...
y-x=6, using y=-2 becomes:
-2-x=6
-x=8
x=-8
So the solution to the system of equations is the point (-8, -2)
You can always check by using these two values in the original equations...
y-x=6, -2--8=6, -2+8=6, 6=6
y+x=-10, -2+(-8)=-10, -2-8=-10, -10=-10