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3 years ago

14.What is the solution of the system?

Mathematics

1 answer:

Archy [21]3 years ago

3 0

Answer:

14. x=-2.5, y = -7

15. x=28 y = -20

Step-by-step explanation:

14. Let's solve this system by elimination. Multiply the first equation by -1.

-1*(4x-y)= -1*(- 3)

-4x +y = 3

Then add this to the second equation.

-4x+y = 3

6x-y= - 8

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

2x = -5

Divide each side by 2

x = -2.5

We still need to find y

-4x+y =3

-4(-2.5) + y =3

10 +y =3

Subtract 10 from each side.

y = 3-10

y = -7

15.I will again use elimination to solve this system, because using substitution will give me fractions which are harder to work with. I will elimiate the y variable. Multiply the first equation by 11

11( 5x+6y)= 11*20

55x+66y = 220

Multiply the second equation by -6

-6(9x+11y)=32*(-6)

-54x-66y = -192

Add the modified equations together.

55x+66y = 220

-54x-66y = -192

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

x = 28

We still need to solve for y

5x+6y = 20

5*28 + 6y =20

140 + 6y = 20

Subtract 140 from each side

6y = -120

Divide by 6

y = -20

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Answer:

La distancia de C con respecto a A es de 197.788 metros.

Step-by-step explanation:

A manera de imagen adjunta construimos una representación del enunciado del problema, la cual representa a un triángulo cuyos tres ángulos son conocidos y la longitud del segmento AB, medida en metros, son conocidos. Por medio de la Ley del Seno podemos calcular la longitud del segmento AC (distancia de C con respecto a A), medida en metros:

$\frac{AB}{\sin C} = \frac{AC}{\sin B}$ (1)

Si sabemos que $B = 57^{\circ}$ , $C = 58^{\circ}$ y $AB = 200\,m$ , entonces la longitud del segmento AC es:

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La distancia de C con respecto a A es de 197.788 metros.

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3 years ago

X+y=20 3x-3y=30 Which values of x and y satisfy this system of equations?

puteri [66]

x +y = 20 ( rewrite as x = 20-y)

3x - 3y =30

3(20-y) - 3y = 30

60-3y - 3y = 30

60-6y = 30

-6y = -30

y = -30 / -6 = 5

x=20-5 = 15

x = 15, y = 5

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4 years ago

If the gradient of the tangent to

Marizza181 [45]

Answer:

Point A(9, 3)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = \sqrt{x}$

$\displaystyle y' = \frac{1}{6}$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = x^{\frac{1}{2}}$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1}$
Simplify: $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x}}$

Step 3: Solve

Find coordinates of A.

x-coordinate

Substitute in y' [Derivative]: $\displaystyle \frac{1}{6} = \frac{1}{2\sqrt{x}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle \frac{1}{3} = \frac{1}{\sqrt{x}}$
[Multiplication Property of Equality] Cross-multiply: $\displaystyle \sqrt{x} = 3$
[Equality Property] Square both sides: $\displaystyle x = 9$