Answer: There are many possibilities such as 108,180,162,126 etc.
Step-by-step explanation:
As it is under 200 the hundreds digit should be 1. And as it should be even, the ones digit should be a multiple of 2. Then finally in the tens place, there should be a number, so that when you add the hundreds digit and ones digit with that number, the sum is 9.
Hope this helps!
It is given that two vertices of square are (0,0) and (4,2).
Now the problem is that you haven't given that whether these two vertices are adjacent vertices or opposite vertices of the square.
1. By Supposing that these two are adjacent vertices of Square
The third vertex will be at (-4,2) which lies in third quadrant.
Suppose the coordinate of fourth vertex be (x,y).
Mid point of line joining (4,2) and (-4,2) is{ [4+(-4)]/2,(2+2)/2} is (0,2).
Mid point of line joining (x,y) and (0,0) is (x/2,y/2).
Since diagonals of square bisect each other,
∵ x/2=0
⇒x=0
and
y/2=2
⇒y=4
So, The Coordinate of fourth vertex is (0,4).
Now coming back to second condition if these are two opposite vertex of Square.
Let the third coordinate be (a,b).
Length of diagonal=
Now,let side of Square be A.
Then length of Diagonal of square =√2 A
⇒√2 A=2√5
⇒A =√10
As third vertex is (a,b).
Using distance formula
a² + b²=10 -------------(1)
(a-4)²+ (b-2)²=10 --------------(2)
Solving expression (1) and (2), we get
⇒a²+ b²=(a-4)² +(b-2)²
⇒2a + b =5
⇒b=5-2a
Putting the value of b in (1),we get
⇒a² +(5-2a)²=10
⇒a²+25+4a²-20a =10
⇒5a²-20a+15=0
⇒a² - 4a + 3=0
Splitting the middle term,we get
⇒(a-3)(a-1)=0
⇒a=3 ∧ a=1
we get b=5-2×1=3 and b=5-2×3=5-6=-1
So,the other vertex are (1,3) and(3,-1).
The other two vertices of the square are (-4, -2) and (-2, 4).
To find the other two vertices of a square with one vertex at (0, 0) and another vertex at (4, 2), you can use the properties of a square, which has equal sides and right angles.
1. First, find the vector from the first vertex (0, 0) to the second vertex (4, 2). This vector represents one side of the square.
Vector = (4 - 0, 2 - 0) = (4, 2)
2. Since the square has equal sides, you can move in the opposite direction of the vector to find the third vertex.
Third Vertex = (0, 0) - (4, 2) = (-4, -2)
3. Now, to find the fourth vertex, you can rotate the vector by 90 degrees counterclockwise. To do this, swap the x and y components and negate the new x component:
Fourth Vertex = (-2, 4)
So, the other two vertices of the square are (-4, -2) and (-2, 4).
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10 and 11
√116 = 10.77 → ( to 2 decimal places )
hence 10 < √116 < 11
O Both the sum and difference have a degree of 7.
The sum has a degree of 6, but the difference has a degree of 7.
The sum has a degree of 7, but the difference has a degree of 6.
The degree of the sum and difference of the polynomials are 6 and 7 respectively.
The degree of the sum and difference of the polynomials.
3x⁵y - 2x³y⁴ - 7xy³ and -8x⁵y + 2x³y⁴ + xy³
The degree of sum = monomial with the highest power = 5 + 1 = 6
The degree of difference = monomial with the highest power = 3 + 4 = 7
Therefore, the degree of the sum and difference of the polynomials are 6 and 7 respectively.
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Answer:
It's C I just did it
Step-by-step explanation: