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brilliants [131]

3 years ago

6

.Resolva as equações, sendo U = Q (separe as respostas por vírgulas). A) 3y - 2 = - y + 4 B) 5 - 4x = x + 3 + 2x C) - 12 - 9x =

- 4 - 12x D) - x - 6 + 3x = -2x + 8x + 15

1 answer:

Oxana [17]3 years ago

5 0

Answer:

$(A) y=1.5$

$(B)x=\dfrac27$

$(C)x=2\dfrac23$

$(D)x=-5.25$

Step-by-step explanation:

(A)

$3y - 2 = - y + 4\\3y+y=4+2\\4y=6\\y=\dfrac64\\\\y=1.5$

(B)

$5 - 4x = x + 3 + 2x\\5 - 4x = 3 +3x\\5-3=3x+4x\\7x=2\\x=\dfrac27$

(C)

$-12 - 9x = - 4 - 12x\\-9x+12x=-4+12\\3x=8\\x=\dfrac83\\\\x=2\dfrac23$

(D) $- x - 6 + 3x = -2x + 8x + 15$

$2x-6 = 6x + 15\\2x-6x=15+6\\-4x=21\\x=\dfrac{21}{-4} \\\\x=-5.25$

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Answer:

$y=-\sqrt{3}x+2$

Step-by-step explanation:

We want to find the equation of a straight line that cuts off an intercept of 2 from the y-axis, and whose perpendicular distance from the origin is 1.

We will let Point M be (x, y). As we know, Point R will be (0, 2) and Point O (the origin) will be (0, 0).

First, we can use the distance formula to determine values for M. The distance formula is given by:

$\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Since we know that the distance between O and M is 1, d=1.

And we will let M(x, y) be (x₂, y₂) and O(0, 0) be (x₁, y₁). So:

$\displaystyle 1=\sqrt{(x-0)^2+(y-0)^2}$

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$1=\sqrt{x^2+y^2}$

We can solve for y. Square both sides:

$1=x^2+y^2$

Rearranging gives:

$y^2=1-x^2$

Take the square root of both sides. Since M is in the first quadrant, we only need to worry about the positive case. Therefore:

$y=\sqrt{1-x^2}$

So, Point M is now given by (we substitute the above equation for y):

$M(x,\sqrt{1-x^2})$

We know that Segment OM is perpendicular to Line RM.

Therefore, their <em>slopes will be negative reciprocals</em> of each other.

So, let’s find the slope of each segment/line. We will use the slope formula given by:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Segment OM:

For OM, we have two points: O(0, 0) and M(x, √(1-x²)). So, the slope will be:

$\displaystyle m_{OM}=\frac{\sqrt{1-x^2}-0}{x-0}=\frac{\sqrt{1-x^2}}{x}$

Line RM:

For RM, we have the two points R(0, 2) and M(x, √(1-x²)). So, the slope will be:

$\displaystyle m_{RM}=\frac{\sqrt{1-x^2}-2}{x-0}=\frac{\sqrt{1-x^2}-2}{x}$

Since their slopes are negative reciprocals of each other, this means that:

$m_{OM}=-(m_{RM})^{-1}$

Substitute:

$\displaystyle \frac{\sqrt{1-x^2}}{x}=-\Big(\frac{\sqrt{1-x^2}-2}{x}\Big)^{-1}$

Now, we can solve for x. Simplify:

$\displaystyle \frac{\sqrt{1-x^2}}{x}=\frac{x}{2-\sqrt{1-x^2}}$

Cross-multiply:

$x(x)=\sqrt{1-x^2}(2-\sqrt{1-x^2})$

Distribute:

$x^2=2\sqrt{1-x^2}-(\sqrt{1-x^2})^2$

Simplify:

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Distribute:

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So:

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Adding 1 and then dividing by 2 yields:

$\displaystyle \frac{1}{2}=\sqrt{1-x^2}$

Then:

$\displaystyle \frac{1}{4}=1-x^2$

Therefore, the value of x is:

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Then, Point M will be:

$\begin{aligned} \displaystyle M(x,\sqrt{1-x^2})&=M(\frac{\sqrt{3}}{2}, \sqrt{1-\Big(\frac{\sqrt{3}}{2}\Big)^2)}\\M&=(\frac{\sqrt3}{2},\frac{1}{2})\end{aligned}$

Therefore, the slope of Line RM will be:

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And since we know that R is (0, 2), R is the y-intercept of RM. Then, using the slope-intercept form:

$y=mx+b$

We can see that the equation of Line RM is:

$y=-\sqrt{3}x+2$

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