Given system of equations:
2y=4x+20 Eq 1
4y=6x+24 Eq 2
Divide both sides of equation 1 by 2.
y=2x+10
Substitute value of y in equation 2.
4(2x+10)=6x+24
8x+40=6x+24
2x+40=24
2x=-16
x=-8
Put value of x in y=2x+10
y=2(-8) + 10
y=-16+10
y=-6
Answer : (-8,-6)
When rounded off then it comes
382.2-297.3=84.9
Please, share just ONE problem at a time. Thanks.
<span>Solve 2x^2-12x+20=0:
Simplify this by dividing each term by 2: x^2 - 6x + 10 = 0
Identify a, b and c: a=1, b=-6 and c=10. Then b^2=36.
Write out the solutions using the quadratic formula:
6 plus or minus sqrt(36-40)
x = ---------------------------------------
2
sqrt(36-40) = sqrt(-4) = plus or minus i2
Then:
6 plus or minus i2
x = --------------------------- (answer)
2</span>
Step-by-step explanation:
The hypotenuse of a right triangle is always the side opposite the right angle. It is the longest side in a right triangle. The other two sides are called the opposite and adjacent sides.
<h2>BRAINILIEST PLEASE </h2>
Here you go!!! Hope this helps
