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3 years ago

I really need help on this question and always post it but never get a response =(

Mathematics

1 answer:

scoray [572]3 years ago

6 0

If both ends of a polynomial both go up or both go down, then the highest power is even
if one end of a polynomial goes up while the other goes down, or vice versa, then the highest power is odd

a positive leading coefficient will result in the right end going upwards
a negative leading coefficient will result in the left end going downwards

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Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

5 0

3 years ago

I need help on these ASAP THX

Kisachek [45]

Answer: The third one or C

4 0

3 years ago

There are 15 roses in a vase of 8 flowers. The rest are daisies.

Strike441 [17]

what is question?

Step-by-step explanation:

g cm n bff e adj k

6 0

3 years ago

Find the slope of the line that passes through (2,2) and (-4,10)

Semmy [17]

Answer:

8/-6 or 4/-3 (either will work, their both the same)

Step-by-step explanation:

Use the formula y2-y1/x2-x1 (For example with this problem, 10-2/-4-2)

Solve it and simplify if you want or it it's needed.

You should get 8/-6 or simplified, 4/-3

5 0

4 years ago

Write the main formulas in perimeter and area in 7ty grade

nadya68 [22]

Answer:

Triangles and Quadrilaterals

P = a+b+c . . . . triangle with sides a, b, c
P = a+b+c+d . . . . general quadrilateral with side lengths a, b, c, d
P = 2(L+W) . . . . rectangle with length L and width W
A = 1/2bh . . . . triangle with base b and height h
A = (1/2)ab·sin(C) . . . . triangle given sides a, b and included angle C
A = √(s(s-a)(s-b)(s-c)) . . . . triangle with sides a, b, c, with s=P/2. "Heron's formula"
A = bh . . . . parallelogram with base b and height h. Includes rectangle and square.
A = ab·sin(θ) . . . . parallelogram with adjacent sides a, b, and included angle θ
A = 1/2(b1+b2)h . . . . trapezoid with parallel bases b1, b2 and height h
A = √((s-a)(s-b)(s-c)(s-d)) . . . . area of cyclic quadrilateral with sides a, b, c, d and s=P/2. "Brahmagupta's formula"

Circles

C = πd = 2πr . . . . circumference of a circle of radius r or diameter d
A = πr² . . . . area of a circle of radius r
A = (1/2)r²θ . . . . area of a circular sector with radius r and central angle θ radians
A = 1/2rs . . . . area of a circular sector with radius r and arc length s

Solids (3-dimensional objects)

A = 2(LW +LH +WH) = 2(LW +H(L+W)) . . . . surface area of a rectangular prism of length L, width W, height H
A = 4πr² . . . . area of a sphere of radius r

Step-by-step explanation:

Perimeter

The perimeter is the sum of the lengths of the sides of a plane figure. When the sides are the same length, multiplication can take the place of repeated addition. Of course, opposite sides of a parallelogram (includes rhombus, rectangle, and square) are the same length, as are adjacent sides of a rhombus (includes square).

The perimeter of a circle is called it circumference. The ratio of the circumference to the diameter is the same for all circles, a transcendental constant named pi (not "pie"), the 16th letter of the Greek alphabet (π). The value of pi is sometimes approximated by 22/7, 3.14, 3.1416, or 355/113. The last fraction is good to 6 decimal places. It has been calculated to several trillion digits.

Area

Fundamentally, the area of a figure is the product of two orthogonal linear dimensions. For odd-shaped figures, the area can be decomposed into smaller pieces that can be added up. (Calculus is used to do this for areas with irregular boundaries.)

The most common figures for which we find areas are triangles, rectangles, and circles. We sometimes need the area of a fraction of a circle, as when we're computing the lateral area of a cone.

It may help you remember the formulas if you notice the similarity of formulas for area of a triangle and area of a circular sector.

Among the formulas above are some that give area when sides and angles are known. The special case of a cyclic quadrilateral (one that can be inscribed in a circle) has its own formula. The similar formula for the area of a triangle from side lengths can be considered to be a special case of the quadrilateral formula where one side is zero.

The formula for area of a trapezoid is somewhat interesting. If the two bases are the same length, the figure is a parallelogram, and the formula matches that for a parallelogram. If one of the bases is zero length, the figure is a triangle, and the formula matches that of a triangle.

In any event, it is useful to note that the area is the product of height and average width. This will be true of any figure — a fact that is used to find the average width in some cases.

_____

Comment on maximum area, minimum perimeter

A polygon will have the largest possible area for a given perimeter, or the smallest possible perimeter for a given area, if it is a regular polygon. For a quadrilateral, the largest area for a given perimeter is that of a square. For a given perimeter, a regular polygon with more sides will have a larger area.

As the number of sides increases toward infinity, the polygon increasingly resembles a circle, which has the largest possible area for a given perimeter.

3 0

4 years ago