Find an equation of the line whose intercepts are twice those of the graph of 5y+2x=10.

2 answers:

8 0

X intercept is where the line crosses the x axis or where y=0
y intercept is where the line crosses the y axis or where x=0

so find x and y intercept

yintercept, x=0
5y=10
y=2
yint is (0,2)

xint, y=0
2x=10
x=5
xint is (5,0)
double

yint is (0,4) and yint is (10,0)

ok, so we do
y=mx+b
m=slope
b=yint
givenyint=4

y=mx+4
when x=10, y=00 because we got (10,0)
0=10m+4
miinus 4 both sides
-4=10m
divide oth sides by 10
-2/5=m

the equation is y=-2/5x+4 or
2/5x+y=4
2x+5y=20

sertanlavr [38]4 years ago

6 0

5y+2x=10y=−25x+2246810−2−4−6−8−10246810−2−4−6−8−10Let's solve for x.5y+2x=10Step 1: Add -5y to both sides.2x+5y+−5y=10+−5y2x=−5y+10Step 2: Divide both sides by 2.2x2=−5y+102x=−52y+5Answer:x=−52y+5

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You are given the dollar value of a product in 2012 and the rate at which the value of the product is expected to change during

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Answer:

$V=r*t+D-2*r$

with

$V=$ dollar value of the product

$r=$ value's rate of change

$t=$ time in years (where t=0 represents 2010)

$D=$ dollar value of product in 2012

Step-by-step explanation:

A linear equation for V in terms of t will have the following form

$V=r*t+C$ (equation 1)

where $r$ will be the rate at which the value of the value of product is expected to change, $V$ is the dollar value of the product at the year $t$ (being $t=0$ for 2010) and $C$ is a constant number which we should be able to find knowing the value of $V$ at some point in time.