Calculus

Simply substitute a 3 in for y. 2(3) + 3 = 6 + 3 = 9.

Note: If this question is from work you have been assigned by a teacher or professor, I suggest you do the work yourself instead of trying to find the answers online. This will help you to better understand the material and become more familiar with it. Of course, if this question was not from homework, you can ignore this message.

There are several definitions. not just one.

Average velocity in a direction = Average displacement (distance) in that direction/time

Instantaneous velocity in a direction = derivative of displacement in that direction with respect to time

Average velocity in a direction = Initial velocity in that direction + Average acceleration in that direction * time

Instantaneous velocity in a direction = Definite integral of acceleration in that direction with respect to time, with initial velocity at t = 0

Then there are others in which time is eliminated.

4260 + 426 ie 4686

"Pascal" is a unit of pressure, not a quantity of air.

The phrase "pascal per minute" is meaningless.

3x3 - 3x2 + 2x - 2 = 3x2(x - 1) + 2(x - 1) = (3x2 + 2)(x - 1)

This is a relatively simple problem, when spread out in front of you. Let's start with:

Y=6X-5

Y=-X+9 or Y=9-X

Since we know that both expressions are equal to Y, they are therefore logically equal to each other as well. When we set them equal to each other, we get:

6X-5=9-X

Next, we get all of the X terms on one side:

7X-5=9

Finally, get all of the constants on the other side.

7X= 14

Now simply solve for X by dividing both sides by the coefficient of X (in this case 7)

X=2

Now that we know what X is equal to, we return to one of our Y= equations. Either one will do, if we did it right they should both equal the same thing. I'll do both here, as a check that we didn't make any arithmetic errors.

Y=6X-5 Y=9-X

Y=6(2)-5 Y=9-(2)

Y=12-5 Y= 7

Y=7

Since both Y's match, we know that we did it properly! In this system of equations, Y is equal to 7 and X is equal to 2

well how you just typed it in the question-supercalafrajalistic-x-pialadotios

-4

True.

But it cannot be "fales", anyway since they is no such word ;D

x2 + 3x - 6 = 0

a = 1, b = 3, c = -6

D = b2 - 4ac = 32 - 4(1)(-6) = 9 + 24 = 33

Since D > 0, the equation has two different real solutions.

1 millimetre = 0.1 centimetres.

Isolate the variable. Get it by itself.

9+11x=89+x

9+11x-1x=89+x-1x Subtraction property of equality

9+10x=89

9+10x-9=89-9 subtraction property of equality

10x=80

x=8 division property of equality.

x would equal sqrt(45) which is ± 6.7082 (to 4 dp)

The answer could be 1260 cubic metres - but only if the three given numbers refer to the lateral dimensions of a cuboid.

The answer depends on the domain for x. For example, if the domain is x = 7, then the range is 55. If the domain is all Real numbers, then the range is y >= 6.

18x +12y = 0

There are several solutions....what are the choices you have?

Guessing one of them is (1,-4) or possibly (-1,-5).

--Algebra 2 Teacher

Yes

only when x is equal to 2

"Grading on a curve" refers to a method of assigning grades that achieves a predetermined statistical distribution, regardless of the absolute level of performance.

"The curve" refers to the Normal Distribution curve, which is the usual distribution of test scores in practice.

For example, giving the top 10% of the class A's, even if no one gets more than half of the exam credit, or failing the bottom 20% of the class, even if everyone scores above 90%.

An advantage of grading on the curve is that it tends to eliminate bias in grades caused by the difficulty of the material or the quality of the instruction. It has the disadvantage that it does not reflect student mastery in absolute terms. It also discourages students from assisting each other with learning the subject.

The degree is equal to the highest exponent on a variable, which is 2.

(2x5) x (55x56) ÷ 6 = 10 x 55 x 56/6 = 10 x 55 x 28/3 = 15400/3 = 5133.3333

It is the required answer.

In general, infinity to the zero is undefined ------------ Seconded. You'll have to figure out some algebra tricks to (re)move the exponent before continuing the evaluation.

This problem is an algebra-based problem using applications of oscillations. The rod is a length L with the axis of rotation in the center.

I found the answer to be that T = PIE x Sq root of (2L(m1 + m2) / (m1 - m2)g)