Answer to Question #179416 in Algebra for kaka

Question

Solve the following; a) If the cost for 6 oranges is 60 rupees, what will be the cost of 6 ¾ dozen oranges? b) How many miles are there between two cities, if the distance is represented by a 4.2 inch line on a map having a scale of 1 inch to 38 miles? c) |3𝑥 − 5| = |−7 − 8𝑥| d) 𝑥 2 + 4𝑥 − 21 , using factoring e) Find the midpoint and the distance of the line segment connecting the points (-3,-7 and 5, -22)

Answer 1

a)

cost of 6 oranges is 60 rupees

cost of 1 orange is $\frac{60}{6}=10$ rupees

1 dozen=12, therefore the cost of $6\frac{3}{4}$ dozen oranges is

$10⋅6\frac{3}{4}⋅12=10⋅\frac{27}{4}⋅12=10⋅27⋅3=810$ rupees

b)

$4.2⋅38=159.60$ miles

c)

$|3x-5|=|-7-8x|$

1) $3x-5=-7-8x$

$3x+8x=5-7$

$11x=-2$

$x=-\frac{2}{11}$

2) $3x-5=-(-7-8x)$

$3x-5=7+8x$

$3x-8x=5+7$

$-5x=12$

$x=-\frac{12}{5}$

3) $-(3x-5)=-7-8x$

$-3x+5=-7-8x$

$-3x+8x=-5-7$

$5x=-12$

$x=-\frac{12}{5}$

4) $-(3x-5)=-(-7-8x)$

$3x+5=7+8x$

$-3x-8x=-5+7$

$-11x=2$

$x=-\frac{2}{11}$

d)

$x^2+4x-21=0$

$x^2+7x-3x-21=0$

$x(x+7)-3(x+7)=0$

$(x-3)(x+7)=0$

$x_1=3$ and $x_2=-7$

e)

midpoint is:

$(\frac{(x_2+x_1)}{2}, \frac{(y_2+y_1)}{2})=(\frac{5-3}{2}, \frac{-22-7}{2})=(1, -14.5)$