Can you find the point of intersection of the lines with equations y equals 7-3x and y equals 2x-3?
Yes I can. Where the lines intersect, (7 - 3x) = (2x - 3). Add (3 + 3x) to both sides of this equation, giving: 10 = 5x or x = 2. Plug x=2 back into either one of the given equations to find 'Y'. If we've done our work correctly, we should get the same 'Y' regardless of which equation we plug x=2 into. And in fact we do: From the first one: Y = 7 - 3x = 7 - (3 times 2) = 7 - 6 = 1 From the second one: Y = 2x - 3 = (2 times 2) - 3 = 4 - 3 = 1 So the two lines intersect at the point ( 2, 1 ).
You need two sets to have an intersection. If you have two sets, call them R and S,
then their intersection is the set T that contains all elements of R that also belong to S OR all elements of S and also belong to R...it's the same thing.
Why you use simpson's method for integrating it can be done by simple integration method also?
Sometimes a function can't be integrated by regular methods, or just an estimate for the integral is needed. It's silly to use Simpson's Rule for a function such as cos(x) that has an elementary antiderivative, but something like e^(-x^2) doesn't. So, you can approximate the definite integral using Simpson's Rule.
Discriminant of 7x to second power minus x plus 1 equals 0?
The discriminant in the quadratic equation is the term B2-4AC. This is the part under the square root sign, with the full form being (-B +/- SquareRoot (B2-4AC)) / 2A.
If the original poloynomial is 7x2 - x + 1 = 0, then A=7, B=-1, and C=1, so the discriminant is -27. Since this is negative, there are two complex conjugate roots.
14x - 2 = x + 6
add 2 to both sides
14x = x +8
subtract x from both sides
13x = 8
x = 8/13
∫ f'(x)/√[f(x)2 + a] dx = ln[f(x) + √(f(x)2 + a)] + C
C is the constant of integration.
integral sin(3 x) cos(5 x) dx = 1/16 (8 cos^2(x)-cos(8 x))+C
Which ordered pair is a point on the graph y equals 3x plus 1?
There are many different ordered pairs for this. To figure it out, make up a value for x. Then plug it into the equation and solve to find y. You can use any number. For example, if x=2, then your equation would be 3(2)+1. Solve that and you get 7 for the answer (y). if x=2, then y=7 so one of the ordered pairs would be (2,7).
How do you solve this 8 5x -6x 30?
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I can see no equation nor inequality.
X2 plus 3x plus 2 over x plus 2 Identify any restricted values on the domain. steps please?
(x2+ 3x + 2) / (x + 2)
= (x + 1)(x + 2) / (x + 2)
= x + 1; but with the following important proviso:
(x + 2) can not equal zero; otherwise we are dividing by zero.
Thus,
(x2+ 3x + 2) / (x + 2) = x + 1, if x ≠ -2; and
(x2+ 3x + 2) / (x + 2) is undefined, if x = -2.
kg*m/s2
This is the units of force in SI. It is not the dimensions. The dimensions are below. -------
Mass * Length * Time-2
Thus, "[M][L][T -2]" are the dimensions of force.
See reference below or more information.
Gray, Costanzo, & Plesha, Engineering Mechanics, Dynamics,McGraw-Hill, 2009, ISBN: 0077-27554-3What is the integral of position?
Physically, it's pretty meaningless. You'll get units of distance times time, so standing 10 metres away from a reference point for two seconds. will give you 20 distance-time's. You can see how this is a fairly meaningless measure.
What is the solution of the equation x2 plus 8x equals 4?
It will have two solutions because its a quadratic equation:
x = -8.472135955 or x = 0.472135955
Program to find the sum of the series as 1 plus x plus x2 plus x3 plus?
The idea is to use a loop. To reduce the additional effort (and innacuracy) of power calculations, you can do repeated multiplication, as part of the loop. For example, in Java:
double sum = 1;
double xpower = 1.0;
for (int i = 1; i <= 100; i++)
{
xpower *= x;
sum += xPower;
}
The idea is to use a loop. To reduce the additional effort (and innacuracy) of power calculations, you can do repeated multiplication, as part of the loop. For example, in Java:
double sum = 1;
double xpower = 1.0;
for (int i = 1; i <= 100; i++)
{
xpower *= x;
sum += xPower;
}
The idea is to use a loop. To reduce the additional effort (and innacuracy) of power calculations, you can do repeated multiplication, as part of the loop. For example, in Java:
double sum = 1;
double xpower = 1.0;
for (int i = 1; i <= 100; i++)
{
xpower *= x;
sum += xPower;
}
The idea is to use a loop. To reduce the additional effort (and innacuracy) of power calculations, you can do repeated multiplication, as part of the loop. For example, in Java:
double sum = 1;
double xpower = 1.0;
for (int i = 1; i <= 100; i++)
{
xpower *= x;
sum += xPower;
}
How do you graph y equals 40x plus 50?
The graph is a straight line, with a slope of 40, passing through the point y=50 on the y-axis.
Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution.