Since it's so nicely grouped, we can work with it! For the equation to equal 0, x=0, 3, or -1 (since x-3 and x+1 equal 0 when plugged in with 3 and -1 respectively). All we have to do is plug in numbers before, between, and after these numbers and apply it to the rest of them. Since -1 is the smallest number of the group, we can plug in a number below that (for this example, -5) and plug it in to get -8*-5*-4= something negative since it contains an odd number of negative numbers. Therefore, anything less than 1 is negative. For between -1 and 0, we get x=-0.5 equals -0.5*-3.5*0.5=something positive (since it has an even amount of negative numbers), proving that everything between -1 and 0 here is positive. For something between 0 and 3, we can plug 1 in to get 1*-2*2= something negative. Do you see a pattern here? It's negative, then positive, etc.. Therefore, if the number is greater than 3 it is positive. Reviewing a bit, we can see that (-inf, -1) is negative as well as (0,3), making the interval notation (-inf, -1) U (0, 3) since when you plug -1, 0, and 3 in it is 0, not less than 0!
Answer:
x = -1 ± √109
Step-by-step explanation:
2x • 3x + (2 • 3)x + 6x = 648
According to PEMDAS (parentheses/exponents | multiplication/division | addition/subtraction), we should solve the parentheses first.
(2 • 3) = 6
Now we have:
2x • 3x + (6)x + 6x = 648
Now let's multiply.
2x • 3x = 6x²
6 • x = 6x
Now we have.
6x² + 6x + 6x = 648
Combine like terms.
6x² + 12x = 648
Let's factor out a 6.
6(x² + 2x) = 648
Divide both sides by 6.
x² + 2x = 108
Let's use completing the square.
Our equation is in a² + bx = c form.
Divide b by 2.
2/2 = 1
Then square it.
1² = 1
Add 1 to both sides.
x² + 2x + 1 = 108 + 1
Simplify.
x² + 2x + 1 = 109
Now we want to factor the left side. A shortcut is just to use b/2.
(x + 1)² = 109
Take the square root of both sides.
x + 1 = ±√109
The square root is as simplified as possible.
Subtract 1 from both sides.
x = -1 ± √109
Hope this helps!
Answer:
C. 1,620 in^3
Step-by-step explanation:
Thank you!
Answer:
the confidence interval for the true weight of the specimen is;
4.1593 ≤ μ ≤ 4.1611
Step-by-step explanation:
We are given;
Standard deviation; σ = 0.001
Sample size; n = 8
Average weight; x¯ = 4.1602
We are given a 99% confidence interval and the z-value at this interval is 2.576
Formula for confidence interval is;
CI = x¯ ± (zσ/√n)
Plugging in the relevant values, we have;
CI = 4.1602 ± (2.576 × 0.001/√8)
CI = 4.1602 ± 0.000911
CI = (4.1602 - 0.000911), (4.1602 + 0.000911)
CI ≈ (4.1593, 4.1611)
Thus, the confidence interval for the true weight will be expressed as;
4.1593 ≤ μ ≤ 4.1611
Where μ represents the true weight
2^4x-2^3x-3
2^3=2x2x2
Which equals 8
