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3 years ago

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In Triangle ABC, the length of Segment AB is equal to the length of segment AC. The length of segment BC is 4 less than twice th

e length of segment AB. If the perimeter of Triangle ABC is 52, find the length of segment AB.

1 answer:

seraphim [82]3 years ago

7 0

Step-by-step explanation:

Let x be the length of segment AB.

Then the length of segment BC is (2x - 4).

The length of segment AC is x.

We know that x + (2x - 4) + x = 52.

Therefore 4x - 4 = 52, 4x = 56, x = 14.

Hence the length of segment AB is 14.

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Select and use the most direct method to solve 2x(x + 1.5) = -1. Describe and justify the methods you used to solve the quadrati

miskamm [114]

Step-by-step explanation:

Given that,

A quadratic equation,

2x(x + 1.5) = -1

We need to solve the quadratic equation. Firstly we need to simplify the above equation to form it as $ax^2+bx+c=0$ .

So,

$2x^2+3x=-1\\\\2x^2+3x+1=0$

Here, a = 2, b = 3 and c = 1

The roots of the given equation can be given by :

$x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

Putting all the values we get :

$x=\dfrac{-b\pm \sqrt{b^2+4ac} }{2a}, \dfrac{-b\pm \sqrt{b^2-4ac} }{2a}\\\\x=\dfrac{-3\pm \sqrt{3^2+4(2)(1)} }{2(2)}, \dfrac{-3\pm \sqrt{3^2-4(2)(1)} }{2(2)}\\\\x=\dfrac{-3+1}{4}, \dfrac{-3-1}{4}\\\\x=-\dfrac{1}{2}, -1$

So, the roots of the given equation is -1/2 and -1.

5 0

3 years ago

Consider the circle of radius 5 centered at (0, 0). Find an equation of the line tangent to the circle at the point (3, 4) in sl

Wittaler [7]

Answer:

$\displaystyle y= -\frac{3}{4} x + \frac{25}{4}$ .

Step-by-step explanation:

The equation of a circle of radius $5$ centered at $(0,0)$ is:

$x^{2} + y^{2} = 5^{2}$ .

$x^{2} + y^{2} = 25$ .

Differentiate implicitly with respect to $x$ to find the slope of tangents to this circle.

$\displaystyle \frac{d}{dx}[x^{2} + y^{2}] = \frac{d}{dx}[25]$

$\displaystyle \frac{d}{dx}(x^{2}) + \frac{d}{dx}(y^{2}) = 0$ .

Apply the power rule and the chain rule. Treat $y$ as a function of $x$ , $f(x)$ .

$\displaystyle \frac{d}{dx}(x^{2}) + \frac{d}{dx}(f(x))^{2} = 0$ .

$\displaystyle \frac{d}{dx}(2x) + \frac{d}{dx}(2f(x)\cdot f^{\prime}(x)) = 0$ .

That is:

$\displaystyle \frac{d}{dx}(2x) + \frac{d}{dx}\left(2y \cdot \frac{dy}{dx}\right) = 0$ .

Solve this equation for $\displaystyle \frac{dy}{dx}$ :

$\displaystyle \frac{dy}{dx} = -\frac{x}{y}$ .

The slope of the tangent to this circle at point $(3, 4)$ will thus equal

$\displaystyle \frac{dy}{dx} = -\frac{3}{4}$ .

Apply the slope-point of a line in a cartesian plane:

$y - y_0 = m(x - x_0)$ , where

$m$ is the gradient of this line, and
$(x_0, y_0)$ are the coordinates of a point on that line.

For the tangent line in this question:

$\displaystyle m = -\frac{3}{4}$ ,
$(x_0, y_0) = (3, 4)$ .

The equation of this tangent line will thus be:

$\displaystyle y - 4 = -\frac{3}{4} (x - 3)$ .

That simplifies to

$\displaystyle y= -\frac{3}{4} x + \frac{25}{4}$ .

3 0

3 years ago

PLS HELP ASAP WILL GIVE BRAINLIEST AND 100 POINTS

konstantin123 [22]

<h2><u>Part A</u>: He started with 4 miles. </h2>

<u>Explanation Part A</u>: 51.664 = (x((1.2⁷)-1)) / 0.2

<h2>
<u>Part B</u>: The equation is 51.664 = (x(1.2⁷)-1)/0.2</h2>

<u>Explanation</u>: n = 7 and on day 7 he traveled 51.664 so Sn = 51.664.

We get the second part of the equation because each day is increased by 20% so we have r = (1.2x) / x = 1.2

So our top half will be x multiplied by ((1.2)⁷-1) because it's the 7th day and we are increasing by 1.2x - 1.

Our bottom half is 1.2 - 1 which equals 0.2

<h2><u>Part C</u> : 103.83 miles</h2>

<u>Explanation: </u> We now have x which is 4 so we plug x into the equation above and remove the 51.664. We also change the 1.2⁷ to 1.2¹⁰ because it's the 10th day not the 7th.

Our equation now looks like (4(1.2¹⁰)-1) / 0.2 which is equal to 103.83 miles.

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2 years ago

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Dmitrij [34]

Answer:

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Step-by-step explanation:

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2 years ago

Solve question 20 (a) and (b)<br>

RUDIKE [14]

Answer:

You first have to read on the coordinates where the curve and the straight line meet or cuts across

a.) 53

b.) 13

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2 years ago