B) two real roots is the answer because it is bigger than 0.
The answer is x = 10, y = 10.
Step 1: rearrange the second equation for y.
Step 2: substitute y from the second equation into the first equation.
Step 3. Calculate y.
Step 1.
<span>The second equation is: 6x + 3y = 90
Divide both sides of the equation by 3:
(6x + 3y)/3 = 90/3
6x/3 + 3y/3 = 30
2x + y = 30
Rearrange the equation:
y = 30 - 2x
Step 2.
</span>Substitute y from the second equation (y = 30 - 2x) into the first equation:
<span>15x + 9y = 240
15x + 9(30 - 2x) = 240
15x + 270 - 18x = 240
15x - 18x = 240 - 270
-3x = -30
x = -30/-3
x = 10
Step 3.
Since </span>y = 30 - 2x and x = 10, then:
y = 30 - 2 * 10
y = 30 - 20
y = 10
Answer:
y = -2x + 16.
Step-by-step explanation:
The slope of the perpendicular line = -1 / slope of the given line
= -1 / 1/2 = -2.
Using the point slope form of the equation of a straight line:
y - y1 = m (x - x1)
y - 8 = -2(x - 4)
y - 8 = -2x + 8
y = -2x + 16 is the required equation.
Answer:
1. Technically 2, but might be 0 in your teacher's opinion.
2. 1
Step-by-step explanation:
Solving problem one.
So I don't know if you have learned about imaginary numbers, but if you have, then you would end up with two answers if you plugged in the quadratic formula.
If you haven't learned about imaginary numbers, then I would say your best option would be to write 'No real solution' since there are technically 2 solutions.
Solving problem two.
Turns out this quadratic has a special property and it's actually a square of one equation. You can find out by just factoring the equation.
It's (3x-2)^2. Since it's squared, that means that only 2/3 would work as x in this equation.
Answer:
h(-3) = -5
Step-by-step explanation:
h(-3) = -I 2-(-3) I
Two negatives makes a positive: -(-3) = +3 or 3
h(-3) = -I 2+3 I
h(-3) = -I 5 I
*There is a negative that lies outside of the absolute value lines, which makes everything positive.
Making sure the 5 is positive, remove the absolute value lines.
h(-3) = - +5
The negative sign is still there.
A negative and a positive makes a negative.
h(-3) = -5
