X intercept is when the function (y) = 0.
0 = 4 - (x+3)^2
(x+3)(x+3) = 4
x^2 + 6x + 9 = 4
x^2 + 6x + 5 = 0
To factor, we find what multiplies to 5 and adds to 6. In this case, it’s 5 and 1.
(x+5)(x+1) = 0
x values of the intercepts are -5 and -1.
The coordinates are (-5,0) and (-1,0).
Answer:
1/12
THIS DESERVES BRAINLISTTT
Step-by-step explanation:
There are 12 possible outcomes, 6 for the die for each of the 2 for the coin. Only one comprises a 5 and a tail so the probability, assuming a fair coin and die, is 1/12.
All you need to do is use the formula to solve. What I do is I plug in the information that I know, and then I just solve. In this case it would be:
V= (4)
- pie(3.14)*4.8(cubed)
3
I hope I was a good help, and if you still need help, come to my page and ask the question on specifically what you need help with. :D
1 liter is equal to 2.11338 pints
To convert 100 liters to pints you would do 100 * 2.11338, or move the decimal place two places to the right, so you have 211.338 pints.
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work
EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same
For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y
For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4
Therefore, for the two expressions to be conjugates, we must satisfy the two conditions.
Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the
x²y = -4 ... (I)
Condition 2: Real parts are the same
x² + y = -3 ... (II)
We have a system of equations since both conditions must be satisfied
x²y = -4 ... (I)
x² + y = -3 ... (II)
We can rearrange equation (II) so that we have
y = -3 - x² ... (II)
Substituting into equation (I)
x²y = -4 ... (I)
x²(-3 - x²) = -4
-3x² - x⁴ = -4
x⁴ + 3x² - 4 = 0
(x² + 4)(x² - 1) = 0
(x² + 4)(x-1)(x+1) = 0
Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.
Solve for y:
y = -3 - x² ... (II)
y = -3 - (±1)²
y = -3 - 1
y = -4
So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:
-3 + ix²y
= -3 + i(±1)²(-4)
= -3 - 4i
x² + y + 4i
= (±1)² - 4 + 4i
= 1 - 4 + 4i
= -3 + 4i
They result in conjugates
