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2 years ago

How do I find x and y?

Mathematics

1 answer:

Andreas93 [3]2 years ago

3 0

First, we solve the determinants.

3 4
D= =(3*3)-(4*5)=9-20=-11
   5 3

   -2 4
Dx= =(-2*3)-(4*4)=-6-16=-22
   4 3


   3 -2
Dy= =(3*4)-(-2*5)=12+10=22
   5    4

Now, we calculate "x" and "y":

x=Dx/D=-22/-11=2
y=Dy/D=22/-11=-2

Solution: x=2; y=-2.

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kipiarov [429]

Answer:

Jumper, y = 172

Shirt, x = - 105

Step-by-step explanation:

Let :

Shirt = x

Jumper = y

5x + 4y = 163 - - (1)

3x + 2y = 29 - - - (2)

Multiply (1) by 3 and (2) by 5

15x + 12y = 489 - - - (3)

15x + 10y = 145 - - - (4)

Subtract :

12y - 10y = 489 - 145

2y = 344

y = 344 / 2

y = 172

Put y = 172 in (3)

15x + 12(172) = 489

15x + 2064 = 489

15x = 489 - 2064

15x = - 1575

x = - 105

The values ;

Jumper, y = 172

Shirt, x = - 105

We weren't supposed to obtain a negative value, kindly check if the parameters given are correct.

5 0

2 years ago

ABCD is a rectangle. The measure of angle BCE is 4x-23 and the angle measure of angle DCE is 5+5x. Find the measure of angle BEC

Temka [501]

<h2>Explanation:</h2><h2></h2>

The diagram for this problem is shown below. Point E is the intersection of the two diagonals and we know that:

$\angle BCE=\beta \\ \\ \angle DCE=\theta$

Every internal angle of any rectangle measures 90 degrees, so:

$\beta +\alpha=90 \\ \\ (4x-23)+(5+5x)=90 \\ \\ \\ Isolating \ x: \\ \\ (4x+5x)+(5-23)=90 \\ \\ 9x-18=90 \\ \\ 9x=108 \\ \\ x=\frac{108}{9}=12$

So:

$\beta=4x-23=4(12)-23=25^{\circ}$

So the measure of angle BEC can be found as follows:

We know that triangle ΔCEB is an isosceles triangle because the diagonals of any rectangle measure the same. So

$\angle EBC=\beta$

The sum of internal angles of any triangle add up to 180 degrees, so:

$\beta + \beta+\angle BEC=180^{\circ} \\ \\ 2\beta+\angle BEC=180^{\circ} \\ \\ \angle BEC=180^{\circ}-2\beta \\ \\ \angle BEC=180^{\circ}-2(25^{\circ}) \\ \\ \angle BEC=180^{\circ}-50^{\circ} \\ \\ \boxed{\angle BEC=130^{\circ}}}$