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2 years ago

Solve the system of equations - 6x + 5y = 0 and -5x + 5y = -5 by combining

Mathematics

1 answer:

Anna [14]2 years ago

7 0

9514 1404 393

Answer:

(x, y) = (-5, -6)

Step-by-step explanation:

The coefficients of y are the same, so we can eliminate y by subtracting one equation from the other. Here, we choose to subtract the first from the second, so that the coefficient of x ends up positive.

(-5x +5y) -(-6x +5y) = (-5) -(0)

x = -5 . . . . . . simplify

-6(-5) +5y = 0 . . . . . . substitute x into the first equation

6 +y = 0 . . . . . . . . . . divide by 5

y = -6 . . . . . . subtract 6

The solution to the system of equations is (x, y) = (-5, -6).

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What is the solution to the system? [1] -3x+4y+2z=-3 [2] 2x-4y-z=0 [3] y=3x-13

Monica [59]

Answer: x = 5

y = 2

z = 2

Step-by-step explanation:

-3x+4y+2z=-3 - ,- - - - - - 1

2x-4y-z=0 - - - - - - - - - - 2

y=3x-13 - - - - - - - - - - 3

We will use the method of substitution.

Substituting y = 3x -13 into equation 1 and equation 2, it becomes

-3x+4(3x-13)+2z=-3

-3x + 12x - 52 + 2z=-3+52

9x+2z = 49 - - - - - - -4

2x - 4(3x-13)-z= 0

2x - 12x + 52 - z = 0

-10x- z = -52 - - - - - - --5

Using elimination method to solve equation 4 and equation 5

Multiply equation 5 by 2 and equation 4 by 1

-20x - 2z = -104

9x+2z = 49

Adding both equations,

-11x = -55

x = -55/-11

x = 5

Substituting x = 5 into equation 3

y = 3x - 13

y = 3×5 -13 = 15-13

y = 2

Substituting x = 5 and y = 2 into equation 1,

-3x+4y+2z=-3

-3×5 + 4×2 +2z = -3

-15+8+2z = -3

-7+2z = -3

2z = -3+7

2z = 4

z = 4/2 = 2

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3 years ago

Use the image below to answer the following question. Find the value of sin x° and cos y°. What relationship do the ratios of si

aliya0001 [1]

Answer:

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

To find sin x and cos y, we must first find the length of the hypotenuse (PO) of the given right triangle.

To do this, use Pythagoras' Theorem:

$a^2+b^2=c^2 \quad \textsf{(where a and b are the legs, and c is the hypotenuse)}$

Given:

a = 6
b = 8
c = PO

$\implies 6^2+8^2=PO^2$

$\implies PO^2=100$

$\implies PO=\sqrt{100}$

$\implies PO=10$

Using the found value of the hypotenuse and the trig ratios quoted above:

$\implies \sf \sin(x)=\dfrac{6}{10}=\dfrac{3}{5}$

$\implies \sf \cos(y)=\dfrac{6}{10}=\dfrac{3}{5}$

Therefore, $\sf \sin(x)=\cos(y)$

4 0

1 year ago

Read 2 more answers

Help please its math i really need help

Norma-Jean [14]

Answer:

m<D = 26

Step-by-step explanation:

Exterior angles thm:

83 + 6x - 4 = 21x + 4

6x + 79 = 21x + 4

6x - 21x = 4 - 79

-15x = -75

x = 5

m<D:

6x - 4

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2 years ago

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koban [17]

Set

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The region $R$ is given in polar coordinates by the set

$R=\left\{(r,\theta)\mid2\le r\le3,0\le\theta\le\dfrac\pi2\right\}$

So we have

$\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_2^3r\sin(r^2)\,\mathrm dr\,\mathrm d\theta=\boxed{\frac\pi4(\cos4-\cos9)}$

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3 years ago

What is the sine cosine and tangent of 360 degrees

Mrac [35]

Sine and cosine has period of 2π and tangent has period of π

So basically same question as sin(0), cos(0), and tan(0)

sin(0) = 0

cos(0) = 1

tan(0) = sin(0) / cos(0) = 0 / 1 = 0

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2 years ago