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3 years ago

If segment AB is 4 units, segment AD is 23 units, and segment CD is 12 units, what is the length of segment BC?

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

3 0

Answer:

<h2>C . 7 units</h2>

Step-by-step explanation:

The question is lacking appropriate diagram. Find the diagram to the question attached.

From the diagram, it can be seen that AD = AB+BC+CD

Given the following

AD = 23units

AB = 4units

CD = 12units

To get C=BC, we will substitute the given segments into the expression above as shown;

23 = 4+BC+12

23 = BC + 4 + 12

23 = BC+16

Subtract 16 from both sides

23-16 = BC+16-16

7 = BC

<em>Hence the length of segment BC is 7units.</em>

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$y=-\dfrac{13}{49}x^2$

Step-by-step explanation:

The shape of an arch corresponds to a parabola.

the general equation for a parabola is:

$y=ax^2+bx+c$

we're given three coordinates: (-7,-13),(7,-13) and (0,0)

so we can plug these values in the general equation to make 3 separate equations:

(x,y) = (-7,-13)

$-13=a(-7)^2+b(-7)+c$

$49a-7b+c=-13$

(x,y) = (7,-13)

$-13=a(7)^2+b(7)+c$

$49a+7b+c=-13$

(x,y) = (0,0)

$0=a(0)^2+b(7)+c$

$c=0$

so we have three equations. and we can solve them simultaneously to find the values of a,b, and c.

we've already found c = 0, let's use substitute it to other equations.

$49a-7b+c=-13\quad\Rightarrow\quad49a-7b=-13$

$49a+7b+c=-13\quad\Rightarrow\quad49a+7b=-13$

we can solve these two equation using the elimination method, by simply adding the two equations

$\quad\quad49a-7b=-13\\+\quad49a+7b=-13$

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

$\quad\quad 98a=-26$

$\quad\quad a=-\dfrac{13}{49}$

Now we can plug this value of a in any of the two equations.

$49a-7b=-13$

$49\left(-\dfrac{13}{49}\right)-7b=-13$

$-13-7b=-13$

$-7b=0$

$b=0$

We have the values of a,b, and c. We can plug them in the general equation to find the equation of the arch.

$y=\left(-\dfrac{13}{49}\right)x^2+0x+0$

$y=-\dfrac{13}{49}x^2$

$49y=-13x^2$

This our equation of the arch!

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