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3 years ago

Endpoint: (1,7), midpoint: (-10,10)

Mathematics

1 answer:

algol [13]3 years ago

5 0

The answer is (-21, 13) for The second endpoint.

Let's start by calling the known endpoint L and the unknown K. We'll call the midpoint M. In order to find this, we must first note that to find a midpoint we need to take the average of the endpoints. To do this we add them together and then divide by 2. So, using that, we can write a formula and solve for each part of the k coordinates. We'll start with just x values.

(Kx + Lx)/2 = Mx

(Kx + 1)/2 = -10

Kx + 1 = -20

Kx = -21

And now we do the same thing for y values

(Ky + Ly)/2 = My

(Ky + 7)/2 = 10

Ky + 7 = 20

Ky = 13

This gives us the final point of (-21, 13)

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Which system of equations could be graphed to solve the equation below? log Subscript 0. 5 Baseline x = log Subscript 3 Baseline

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You can use the fact that two expressions in equality can be considered to be equal to a third variable(not used in given context).

The system of equations that could be graphed to solve the equation given is

$y = \log_{0.5}(x)\\\\$
$y = \log_3(2+x)$

<h3>How can we form a system of equations from an equation?</h3>

Suppose the equation be $a = b\\$

Let there is a symbol $c$ such that we have $a = b = c$

It is because a and b are same measure (that is exactly what a = b means)

and we gave another name c to that measure.

Thus, we have

$a = c\\b = c$

in addition to $a = b\\$

<h3>Using the above method to find the system of equations needed</h3>

Since the given equation is $log_{0.5}(x) = log_2(2+x)$

The 2d graphs are usually expressed as $y = f(x)$ on X-Y plane.

Taking the equation's expressions equal to y, we get

$log_{0.5}(x) = log_2(2+x) = y$

or, we get system of equations as

$y = log_{0.5}(x)\\\\y = log_3{(2 + x)}$

Their graph is plotted below. The intersection point of both curves is the solution to the given equation as it satisfies both the equations of the system of equations formed from the given equation.

Thus,

The system of equations that could be graphed to solve the equation given is