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2 years ago

.162q + .035c = 4 Solve for q and c and show work

Mathematics

1 answer:

liberstina [14]2 years ago

7 0

Answer:

q = $\frac{5(-7c+800)}{162}$

c = $\frac{2(-81q + 2000)}{35}$

Step-by-step explanation:

First solve for q:

0.162q + 0.035c = 4

0.162q = 4 - 0.035c (move term to other side)

q = $\frac{4-0.035c}{0.162}$ (divide both sides by 0.162 to get it away from q)

q = $\frac{5(-7c+800)}{162}$ (simplify fraction)

Solve for c:

0.162q + 0.035c = 4

0.035c = 4 - 0.162q (move term to other side)

c = $\frac{4 - 0.162q}{0.035}$ (divide both sides by 0.035 to get it away from c)

c = $\frac{2(-81q + 2000)}{35}$ (simplify fraction)

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3 years ago

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3 years ago

The distance between the two points pictured is d= Vn

katrin [286]

the distance between points is:

d = 7.8 units

d = root ((x2-x1) ^ 2 + (y2-y1) ^ 2)

The ordered pairs are:

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3 years ago

A bridge in the shape of an arch connects two cities separated by a river. The two ends of the bridge are located at (–7, –13) a

sdas [7]

Answer:

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

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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2 years ago