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ELEN [110]
3 years ago
5

Three vertices of parallelogram JKLM are J(1, 4), K(5, 3), and L(6,−3). Find the coordinates of vertex M.

Mathematics
1 answer:
Andrej [43]3 years ago
5 0

Answer:

coordinates of vertex M is (x, y) = (2, -2)

Step-by-step explanation:

Since JKLM is a  parallelogram, this implies that JK parallel to LM and KL parallel to JM. This means that

Slope of JK = slope of LM

\frac{3-4}{5-1} =\frac{-3-y}{6-x} \\y=-\frac{1}{4}x-\frac{3}{2} ....(i)

And

Slope of KL = slope of JM

\frac{3-\left(-3\right)}{5-6}=\frac{4-y}{1-x}\\y=10-6x...(ii)

From equation (i) and (ii) we get

-\frac{1}{4}x-\frac{3}{2} =10-6x

-\frac{23x}{4}=-\frac{23}{2}

-23x=-46

x=2

Put the value of x in equation (ii) we get

y=10-6(2)\\y=-2

So, the coordinates of vertex M is (x, y) = (2, -2).

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What is the distance of KF. round to the nearest tenths ​
snow_tiger [21]

Answer:

its 9.2

Step-by-step explanation:

Well i counted 7 units (x-axis) and 6 units (y axis). 6 squared is 36 and 7 squared is 49. Add them together and get 85. Square root 85 and youll get 9.21954445729. Round it to the nearest tenths and get 9.2

5 0
3 years ago
The length of a rectangle is 2 cm less than 7 times the width. The perimeter is 60 cm. Find the width of the rectangle. HINT: Dr
klemol [59]

Answer:

Step-by-step explanation:

Perimeter of a rectangle = 2L +2W

L is the length

W is the width

If the length of a rectangle is 2 cm less than 7 times the width, then;

L = 7W-2 .... 2

Substitute eqn into the formula for finding the perimeter

P = 2(7W-2)+2W

P = 14W-4+2W

P = 16W-4

If P =60cm

60 = 16W-4

16W = 60+4

16W = 64

W = 64/16

W = 4cm

Hence the width of the rectangle is 4cm

Since l = 7w-

l = 7(4)-2

L = 28-2

L = 26cm

You can proceed and draw a rectangle with the dimension of 4cm by 26cm

3 0
3 years ago
The Futurists, a rock group, are planning a concert tour with performances to be
Marrrta [24]

Answer:

a. 24

b. 12

Step-by-step explanation:

a. You can choose from 4 cities for the first performance, then from 3 cities, then from 2 cities, then the last city.

4 * 3 * 2 * 1 = 24

b. Think of the three performances in California as a unit. Within this unit, you can choose from 3 cities for the first one, then 2 cities for the second one, and finally 1 city for the last one.

3 * 2 * 1 = 6

Now you can have the 3 California performances first, followed by Denver, or Denver first followed by California.

6 * 2 = 12

7 0
3 years ago
Can someone please help me with question 1, I will mark you as a brainest and I'll thank you
Leya [2.2K]
I believe it is 86.4
8 0
3 years ago
Linear Algebra question! Please help!
kozerog [31]

Answers:

  1. false
  2. false
  3. true
  4. false
  5. True

==================================================

Explanation:

Problem 1

This is false because the A and B should swap places. It should be (AB)^{-1} = B^{-1}A^{-1}.

The short proof is to multiply AB with its inverse (AB)^{-1}  and we get: (AB)*(AB)^{-1} = (AB)*(B^{-1}A^{-1}) = A(B*B^{-1})*A^{-1} = A*A^{-1} = I

The fact we get the identity matrix proves that we have the proper order at this point. The swap happens so that B matches up its corresponding inverse B^{-1} and the two cancel each other out.

Keep in mind matrix multiplication is <u>not</u> commutative. So AB is not the same as BA.

-------------------------

Problem 2

This statement is true if and only if AB = BA

(A+B)^2 = (A+B)(A+B)

(A+B)^2 = A(A+B) + B(A+B)

(A+B)^2 = A^2 + AB + BA + B^2

(A+B)^2 = A^2 + 2AB + B^2 ... only works if AB = BA

However, in most general settings, matrix multiplication is <u>not</u> commutative. The order is important when multiplying most two matrices. Only for special circumstances is when AB = BA going to happen. In general,  AB = BA is false which is why statement two breaks down and is false in general.

-------------------------

Problem 3

This statement is true.

If A and B are invertible, then so is AB.

This is because both A^{-1} and B^{-1} are known to exist (otherwise A and B wouldn't be invertible) and we can use the rule mentioned in problem 1. Make sure to swap the terms of course.

Or you can use a determinant argument to prove the claim

det(A*B) = det(A)*det(B)

Since A and B are invertible, their determinants det(A) and det(B) are nonzero which makes the right hand side nonzero. Therefore det(A*B) is nonzero and AB has an inverse.

So if we have two invertible matrices, then their product is also invertible. This idea can be scaled up to include things like A^4*B^3 being also invertible.

If you wanted, you can carefully go through it like this:

  1. If A and B are invertible, then so is AB
  2. If A and AB are invertible, then so is A*AB = A^2B
  3. If A and A^2B are invertible, then so is A*A^2B = A^3B

and so on until you build up to A^4*B^3. Therefore, we can conclude that A^m*B^n is also invertible. Be careful about the order of multiplying the matrices. Something like A*AB is different from AB*A, the first of which is useful while the second is not.

So this is why statement 3 is true.

-------------------------

Problem 4

This is false. Possibly a quick counter-example is to consider these two matrices

A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \text{ and } B = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix}

both of which are invertible since their determinant is nonzero (recall the determinant of a diagonal matrix is simply the product along the diagonal entries). So it's not too hard to show that the determinant of each is 1, and each matrix shown is invertible.

However, adding those two mentioned matrices gets us the 2x2 zero matrix, which is a matrix of nothing but zeros. Clearly the zero matrix has determinant zero and is therefore not invertible.

There are some cases when A+B may be invertible, but it's not true in general.

-------------------------

Problem 5

This is true because each A pairs up with an A^{-1} to cancel out (similar what happened with problem 1). For more info, check out the concept of diagonalization.

5 0
2 years ago
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