X-3y=6
x+y=2
this is an substitution problem
so first you can do is rewrite the problem by subjection one variable
x=3y+6
then substitute this in the other proble
x+y=2
(3y+6)+y=2
4y+6=2
4y=2-6
4y=-4
y=-1
then substitute the no. in the original equation.
x=3y+6
x=3(-1)+6
x=-3+6
x=3
now you got the intercepts and you draw the line and check.
it's in the IV quadrant
You would go to Neptune, Uranus, Saturn, Jupiter, Mars, Earth (to refuel), Mercury, Venus
Answer:
m∠QPM=43°
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
m∠NPQ=m∠MPN+m∠MPQ
we have
m∠NPQ=(9x-25)°
m∠MPN=(4x+12)°
m∠MPQ=(3x-5)°
substitute the given values and solve for x
(9x-25)°=(4x+12)°+(3x-5)°
(9x-25)°=(7x+7)°
9x-7x=25+7
2x=32
x=16
Find the measure of angle QPM
Remember that
m∠QPM=m∠MPQ
m∠MPQ=(3x-5)°
substitute the value of x
m∠MPQ=(3(16)-5)=43°
therefore
m∠QPM=43°
<span>
First, (again lol) you multiply the amount of chocolate chips the recipe calls for by the amount Lee wants to make.</span>
<span />so its 3/4 * 2/3
<span> =</span>
<span>6/12</span>
<span>1/2</span>
<span>Lee will need a fraction of 1/2 of a chip of chocolate chips for the recipe</span>
To find the area of his exclusion zone you would need to understand that a triangle with dimensions of 3, 4, and 5 represent a right triangle.
This means the exclusion zone would be applied to the base and the height of the triangular space.
You would add 2 km to the 3 km, and 2 km to the 4 km to create a new height of 5 km and a new base of 6 km.
Please see the attached picture to understand this.
You will find the area of the total space created by the new triangle and subtract the space represented by the original triangle to find the area of the exclusion zone.
(1/2 x 6 x 5) - (1/2 x 4 x 3)
15 km² -6 km² equals 9 km².
The exclusion space is 9 km².