It is Y.
The tangent (of a curve) is a vector that is tangent (perpendicular to the normal), i.e. the instantaneous velocity of the curve at a specific point. As such, the initial tangent is the initial velocity of the curve at the point where t=0. Stated in other terms, the tangent is the slope of the line at a point. This is expressed (in two dimensions, but applicable to higher dimensions), as the line that has x and y coordinates equal to the point of tangency, and slope equal to the limit of delta y over delta x as delta x (and delta y) approaches zero.
If y = 2x+1 is a tangent line to the circle 5y^2 +5x^2 = 1 then the point of contact is at (-2/5, 1/5) because it has equal roots
the formula for the arc of a triangle is the arc length is equal to the angle times the radius. s=arc length theta=angle made y length of the arc lenth r=radius s=theta times radius
To get the tangent, you first need to find it's slope: y = x2 ∴ dy/dx = 2x So the slope with respect to x is equal to 2x. When x = 2 then, the slope is 4. Now you need the y-coordinate that occurs when x equals 2. We know that y = x2, so when x equals 2, y equals 4. You now have a point on the tangent, as well as it's slope, allowing you to define the line: Recall the definition of a line: Δy = sΔx or: y - y1 = s(x - x1) ∴ y - 4 = 4(x - 2) ∴ y = 4 + 4x - 8 ∴ y = 4x - 4 And that is your tangent.
If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2
If: y = kx+1 is a tangent to the curve y^2 = 8x Then k must equal 2 for the discriminant to equal zero when the given equations are merged together to equal zero.
The tangent (of a curve) is a vector that is tangent (perpendicular to the normal), i.e. the instantaneous velocity of the curve at a specific point. As such, the initial tangent is the initial velocity of the curve at the point where t=0. Stated in other terms, the tangent is the slope of the line at a point. This is expressed (in two dimensions, but applicable to higher dimensions), as the line that has x and y coordinates equal to the point of tangency, and slope equal to the limit of delta y over delta x as delta x (and delta y) approaches zero.
It is a function which maps the tangent ratio - any real value - to an angle in the range (-pi/2, pi/2) radians. Or (-90, 90) degrees.If tan(x) = y then x is the inverse tangent of y.It is also known as "arc tangent", and spreadsheets, such as Excel, use "atan" for this function.Warning:1/tangent = cotangent is the reciprocal, NOT the inverse.
The tangent of an angle theta is defined as sine(theta) divided by cosine(theta). Since the sine and cosine are Y and X on the unit circle, then tangent(theta) is Y divided by X. The tangent of a function at a point is the line going through that point which has slope equal to the first deriviative of the function at that point.
The question seems to be incomplete!
Normally a straight line is a tangent to a curved line but, presumably, that relationship can be reversed. So a tangent to the y axis would be a curve that just touches the y axis but does not cross it - at least, not at the point of tangency.
tan y = 20/15
the formula for the arc of a triangle is the arc length is equal to the angle times the radius. s=arc length theta=angle made y length of the arc lenth r=radius s=theta times radius
If y = 2x+1 is a tangent line to the circle 5y^2 +5x^2 = 1 then the point of contact is at (-2/5, 1/5) because it has equal roots
To get the tangent, you first need to find it's slope: y = x2 ∴ dy/dx = 2x So the slope with respect to x is equal to 2x. When x = 2 then, the slope is 4. Now you need the y-coordinate that occurs when x equals 2. We know that y = x2, so when x equals 2, y equals 4. You now have a point on the tangent, as well as it's slope, allowing you to define the line: Recall the definition of a line: Δy = sΔx or: y - y1 = s(x - x1) ∴ y - 4 = 4(x - 2) ∴ y = 4 + 4x - 8 ∴ y = 4x - 4 And that is your tangent.
If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2
There is not a way to give the answer to this equation. You will have to give me the value of the letters.