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Otrada [13]
3 years ago
6

Use factoring to find the zeros. x2 - 9x – 36 = 0

Mathematics
1 answer:
3241004551 [841]3 years ago
4 0

Answer: x=12, x=-3

Step-by-step explanation: For this factored problem, you see that -36 is negative, as well at -9. This shows you that you need to find two numbers that multiply to get -36, and add up to get -9. Two numbers that both satisfy these requirements are -12 and 3. They multiply to get -36, and add to get -9.

Once you have found your values, you plug them into (x+/- __)(x+/- __), which is (x-12)(x+3).

But since the problem asks you to find the zeroes, you set each factor equal to zero, like this:

x-12=0 and x+3=0

Then solve and you have you answer!

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pickupchik [31]
Give me an operation like the one you are talki ng about and I can solve it
8 0
3 years ago
Hey can you please help me posted picture of question
jenyasd209 [6]

The equation of parabola is x=ay^2. If a is positive and y^2 is always greater than zero or equal to zero, then x is also greater or equal to zero. This means that parabola is determined for non-negative x and for all real y.

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Answer: correct choice is A.

6 0
3 years ago
When the solution of x2 − 9x − 6 is expressed as 9 plus or minus the square root of r, all over 2, what is the value of r? x equ
jarptica [38.1K]

Answer:

r = 105

Step-by-step explanation:

Given quadratic equation:

x² − 9x − 6

Quadratic formula is given by:

(-b ± √b² - 4ac) ÷ 2a

For the equation: x² - 9x - 6

a = 1, b = -9 and c = -6

Substitute in the quadratic formula:

(-b ± √b² - 4ac) ÷ 2a

(-(-9) ± √(-9)² - 4(1*-6)) ÷ 2*1

= 9 ± √105 / 2

The solution of x² − 9x − 6 is expressed as 9 plus or minus the square root of r, all over 2

Therefore, the value of r is 105

4 0
3 years ago
Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist
yuradex [85]

Answer:

0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.

Step-by-step explanation:

Given the differential equation

(t - 5)y' + (ln t)y = 6t

and the condition y(1) = 6

We can rewrite the differential equation by dividing it by (t - 5) as

y' + [(ln t)/(t - 5)]y = 6t/(t - 5)

(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).

6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)

We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).

But the initial condition is y = 6, when t = 1.

The solution to differential equation is certain to exist at (0, 5)

Which implies that

0 < t < 5

is the required interval.

3 0
4 years ago
What is 10 3/4 - 5 1/4=
Temka [501]

you can make it into an improper fraction if you want but there is no need for this problem

steps

1) 10 3/4- 5 1/4

10-5=5

3/4-1/4=2/4 or 1/2

answer:5 1/2 or 5 2/4

(it is better simplified)

8 0
3 years ago
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