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3 years ago

How many solutions does this system have? How do you know? { 3x + y= 2 4 = 12x − 12

Mathematics

1 answer:

lesya [120]3 years ago

5 0

This system has one solution, because both of the equations are linear. The answer is (4/3, -2).

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3 Find the value of a in the equation,(3+4√3)(2-a√3)=-18+2√3

ankoles [38]

Given:

The equation is:

$\left(3+4\sqrt{3}\right)\left(2-a\sqrt{3}\right)=-18+2\sqrt{3}$

To find:

The value of a.

Solution:

We have,

$\left(3+4\sqrt{3}\right)\left(2-a\sqrt{3}\right)=-18+2\sqrt{3}$

On simplification, we get

$(3)(2)+(4\sqrt{3})(2)+(3)(-a\sqrt{3})+(4\sqrt{3})(-a\sqrt{3})=-18+2\sqrt{3}$

$6+8\sqrt{3}-3a\sqrt{3}-4a(3)=-18+2\sqrt{3}$

$6+8\sqrt{3}-3a\sqrt{3}-12a=-18+2\sqrt{3}$

$(6-12a)+(8-3a)\sqrt{3}=-18+2\sqrt{3}$

On comparing both sides, we get

$6-12a=-18$

$-12a=-18-6$

$a=\dfrac{-24}{-12}$

$a=2$

And,

$8-3a=2$

$-3a=2-8$

$a=\dfrac{-6}{-3}$

$a=2$

Therefore, the value of a is 2.

3 0

3 years ago

Question in the image below:

horrorfan [7]

Answer:

x = 5/33 or 0.151515151

Step-by-step explanation:

7(-6x-2) = 8(3x-4)

step 1: simplify the equation.

-42x - 14 = 24x - 4

step 2: isolate the variable (using the balance method).

-42x - 14 + 14 = 24x - 4 + 14

-42x = 24x + 10

-42x - 24x = 24x - 24x + 10

-66x = 10

step 3: solve for x.

x = 10 ÷ -66

x = 5/33

4 0

3 years ago

When your income is more than your expenses, you have _____.

blagie [28]

When your income is more than your expenses, you have surplus

3 0

3 years ago

Read 2 more answers

A 2D vector can have a component equal to zero even when its magnitude is nonzero.

lys-0071 [83]

Answer:

T F F F T F T

Step-by-step explanation:

A 2D vector can have a component equal to zero even when its magnitude is nonzero. TRUE.

it will be actually a 1d vector, but it's still a 2d vector too. this happens when the vector it's aligned to one of the axis.

The direction of a vector can be different in different coordinate systems. FALSE

The direction of a vector is independent of any coordinate system.

A 2D vector can have a magnitude equal to zero even when one of its components it nonzero. FALSE.

because the only way $\sqrt{x^{2} +y^{2} } =0$ is that both x and y are equal to zero

The magnitude of a vector can be different in different coordinate systems. FALSE.

The magnitude stays the same in every coordinate system

It is possible to multiply a vector by a scalar. TRUE.

it changes only it's magnitude.

It is possible to add a scalar to a vector. FALSE.

you can´t sum elements of diferent spaces. R, $R^{2}$ , etc

The components of a vector can be different in different coordinate systems.TRUE.

The components of a vector are just a way for identifying the vector, in a determinate coordinate system, the way i call those components will change as I change the name I call every point in the plane.

5 0

3 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago