(0, -3)(2, -2)
slope = (-2 + 3)/(2-0) = 1/2
b = -3
equation
y = 1/2x - 3
answer is A
<span>y = 1/2x - 3</span>
Well, 50 points total given, 25 points per user and 13 bonus for brainliest
anyway
1.
add them equations, the y's will cancel
x+2y=4
<span>3x-2y=4 +</span>
4x+0y=8
4x=8
divide by 4 both sides
x=2
sub back
x+2y=4
2+2y=4
minus 2
2y=2
divide 2
y=1
x=2
y=1
(x,y)
(2,1) is solution
2.
the solution is where they intersect
multiply 2nd equation by 2 and add to first
4x-14y=6
<span>-4x+14y=-6 +</span>
0x+0y=0
0=0
infinite solutions
that is because they are actually the same line
the solutions are (x,y) such that they satisfy -2x+7y=-3 or 4x-14y=6 (same equaiton)
infinite solutions
3.
multiply first equation by 2 and add to first
4x+2y=-6
<span>1x-2y=-4 +</span>
5x+0y=-10
5x=-10
divide by 5 both sides
x=-2
sub bac
x-2y=-4
-2-2y=-4
add 2
-2y=-2
divide by -2
y=1
x=-2
y=1
(x,y)
(-2,1)
4.
coincident means they are the same line
so
we see that we have to multiply 4 by 2 to get 8
multiply top equation by 2
8x+10y=16
8x+By=C
B=10 and C=16
5.
a. false, either 0, 1, or infinity solutions
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'
b.false
'sometimes' to 'always'
c. true
d. false, change 'sometimes' to 'always'
EC
total cost=150
TC=childC+adultC
TC=3c+5a
150=3c+5a
40 tickets, c+a
40=c+a
the equations are
150=3c+5a and
40=c+a
eliminate
multiply 2nd equaton by -3 and ad to first one
150=3c+5a
<span>-120=-3c-3a +</span>
30=0c+2a
30=2a
divide by 2
15=a
sub back
40=c+a
40=c+15
minus 15
25=c
25 children tickets and 15 adult tickets were sold
ANSWERS:
1.
(2,1) is solution
2.
infinite solutions
3.
(-2,1)
4.
B=10 and C=16
5.
a. false,
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'
b.false
'sometimes' to 'always'
c. true
d. false, change 'sometimes' to 'always'
EC
the equations are
150=3c+5a and
40=c+a
25 children tickets and 15 adult tickets were sold
Answer:
Step-by-step explanation:
4 is coefficient
x is variable
12 is constant
Answer:
B
Step-by-step explanation:
We are given that <em>x</em> and <em>y</em> are functions of time <em>t</em> such that <em>x</em> and <em>y</em> is a constant. So, we can write the following equation:
The rate of change of <em>x</em> and the rate of change of <em>y</em> with respect to time <em>t</em> is simply dx/dt and dy/dt, respectively. So, we will differentiate both sides with respect to <em>t: </em>
<em />![\displaystyle \frac{d}{dt}\Big[x(t)+y(t)\Big]=\frac{d}{dt}[k]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdt%7D%5CBig%5Bx%28t%29%2By%28t%29%5CBig%5D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Bk%5D)
<em />
Remember that the derivative of a constant is always 0. Therefore:
And by subtracting dy/dt from both sides, we acquire:
Hence, our answer is B.
