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2 years ago

If f(x) = x2 + x, find f(-3). -12 3 6

Mathematics

2 answers:

ANEK [815]2 years ago

4 0

Answer:

-9

Step-by-step explanation:

-3(2) + (-3)

-6 + (-3)

-9

makkiz [27]2 years ago

4 0

Answer:

f(-3) = 6

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Step-by-step explanation:

Step 1: Define

f(x) = x² + x

f(-3) is x = -3

Step 2: Evaluate

Substitute: f(-3) = (-3)² + -3
Exponents: f(-3) = 9 - 3
Subtract: f(-3) = 6

And we have our final answer!

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Answer:

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Step-by-step explanation:

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2 years ago

Last Wednesday it snowed 3 1/4 in. On Sunday it snowed 3 4/5 in. On which day did it snow more? How much more?

damaskus [11]

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3 years ago

Read 2 more answers

Calculate the expected value of playing a lottery ticket game where each ticket costs $5, and the chance of having winning ticke

Amanda [17]

Answer:

$4895

Step-by-step explanation:

P(winning) = 1 / 1000000 = 0.000001

P(not winning) = 1 - 1/1000000 = 0.999999

Winning amount = 4,900,000,000 - 5 = 4899999995

Amount lost = - 5

__ X: - 5 _________ 4899999995

P(X) : 0.999999 ___ 0.000001

Expected value E(x) = Σx * p(x)

E(x) = - 5(0.999999) + 4899999995(0.000001)

E(x) = - 4.999995 + 4899.999995

E(x) = 4895

Hence, expected value is 4895

6 0

2 years ago

Which of the following is an equivalent to the expression shown? Hint: factor! 7x−56−28x Group of answer choices 7(x-4-4x) -7(x-

Anestetic [448]

Answer:i think its the second one

Step-by-step explanation:

took the test

3 0

3 years ago

Two airplanes are flying in the air at the same height. Airplane A is flying east at 300mi/h and airplane B is flying north at 2

Ipatiy [6.2K]

Answer:

$\frac{dAB}{dt}=340 mil/h$

Step-by-step explanation:

The change of distance over time of the plain A is 300 mi/hour and 200 mi/hour for plane B. O is the point of the airport.

So, the distance from A to O AO = 90 miles and BO = 120 miles.

Now, we have a right triangle here. We can use the Pythagorean theorem, so the distance between the planes will be:

$AB^{2} =AO^{2}+BO^{2}$ (1)

$AB =\sqrt{AO^{2}+BO^{2}}=$

$AB =\sqrt{90^{2}+120^{2}}=150 miles$

If we take the derivative of the equation (1) we could find the change of the distance between planes.

$2AB\frac{dAB}{dt}=2AO\frac{dAO}{dt}+2BO\frac{dBO}{dt}$

$2*150\frac{dAB}{dt}=2*90*300+2*120*200=102000 mil/h$

$\frac{dAB}{dt}=\frac{102000}{150*2}$

Finally,

$\frac{dAB}{dt}=340 mil/h$

I hope it helps you!

6 0

2 years ago