Hmm try 192.5 I think that’s right
the parallel line is 2x+5y+15=0.
Step-by-step explanation:
ok I hope it will work
soo,
Solution
given,
given parallel line 2x+5y=15
which goes through the point (-10,1)
now,
let 2x+5y=15 be equation no.1
then the line which is parallel to the equation 1st
2 x+5y+k = 0 let it be equation no.2
now the equation no.2 passes through the point (-10,1)
or, 2x+5y+k =0
or, 2*-10+5*1+k= 0
or, -20+5+k= 0
or, -15+k= 0
or, k= 15
putting the value of k in equation no.2 we get,
or, 2x+5y+k=0
or, 2x+5y+15=0
which is a required line.
Answer:
The equation that represents the new path is y=(1/2)x-2
Step-by-step explanation:
step 1
Find the slope of the give line
we have
y=-2x-7
so
the slope m is equal to
m=-2
step 2
Find the slope of the perpendicular line to the given line
Remember that
If two lines are perpendicular, then their slopes are opposite reciprocal of each other
so
we have
m=-2 -----> slope of the given line
therefore
The slope of the perpendicular line is equal to
m=1/2
step 3
With m=1/2 and the point (-2,-3) find the equation of the line
y-y1=m(x-x1)
substitute
y+3=(1/2)(x+2)
y=(1/2)x+1-3
y=(1/2)x-2 -----> equation that represent the new path
The practical domain is all real numbers from 4 to 9, inclusive.
The practical range is all real numbers from 49.6 to 111.6, inclusive.
These are the correct options.
Explanation:
Given function is f(t) = 12.4t
Let us assume that Nate works 'x' hours so 4<x<9
And multiplying the hours with his earnings we get the range.
4*12.40=49.6 and 9*12.40=111.6. Let the amount earned be represented by y
Hence, domain can be represented as 4<x<9 and range can be represented as 49.6 < y < 111.6
There are 18 students in Grade 7, 10 are in band, so 10/18 = 0.56 are in band.
There are 12 students in Grade 8, 10 are in band, so 10/12 = 0.83 are in band.
Students in Grade 8 are more likely to be members of the band.
