Can someone help me with this question I don't know how to subtract a whole number by a fraction

The answer is:

120

Given the arithmetic sequence

3, 5, 7, 9, ..., 21

The first term a = 3 & a common difference d = 5 - 3 = 7 - 5 = ... = 2

If there are n number of terms in the above A.P. then last term l = 21 will be the nth term given as:

l = a + (n - 1)d

21 = 3 +(n − 1)2

n = 10

Hence the sum of given arithmetic progression (A. P.) up to 10 terms is given general formula.

Sn = n/2(a + l)

S10 = 10/2(3 + 21)

= 120

Answer:

Option A - 5 grams

Step-by-step explanation:

9 balls in one side

= 5 balls other side with 20 gram

5 + 4 balls one side

= 5 + 20 grams other side

5 + 4 balls

= 5 balls + 20/4

5 gm of I ball weight 9 * 5 = 45

5 * 5 + 20

= 45

Answer:

b = 5 grams

Step-by-step explanation:

Let b represent the weight of one ball.

Then 9b = 20 grams + 5b.

Solving for b, subtract 5b from both sides, obtaining 4b = 20 grams.

Then b = the weight of one steel ball = 5 grams.

Check:  Does 9(5 grams) = 20 grams + 5(5 grams)?

              Does 45 grams = 45 grams?  YES

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).

What are the other two vertices of a square with one vertex (0, 0) and another vertex (4, 2)?

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).

Learn more on vertices here;

brainly.com/question/1217219

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