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3 years ago

Calculus Help Show that one function is not an antiderivative of another

Mathematics

1 answer:

liraira [26]3 years ago

6 0

Just differentiate G and H :

• dG/dx = d(x ² eˣ )/dx = 2x eˣ + x ² eˣ

so G is not an antiderivative of f.

• dH/dx = d(2x eˣ - 2eˣ ) = (2eˣ + 2x eˣ ) - 2eˣ = 2x eˣ = f(x)

so H is indeed an antiderivative of f.

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The perimeter of a rectangle is 56 inches. The rectangle is 20 inches long. How wide is the rectangle?

Mamont248 [21]

Answer:

The answer is A 8 inches

Step-by-step explanation:

If the perimeter is 56 inches and it is 20 inches long, then you would multiply 20 by two and get 40, then you subtract 40 from 56 and get 16. Finally, you divide 16 by two and get 8. So, A 8 inches will be your answer.

5 0

3 years ago

Rob is making a tent out of canvas in the shape of a square pyramid. Each side of the square base is 6 feet long, and the height

Elena L [17]

The amount of balloons in the tent is 125.075 feet cube. From the given options, approximately, option D is correct

Solution:

Given, Rob is making a tent out of canvas in the shape of a square pyramid. Each side of the square base is 6 feet long, and the height is 6.3 feet.

Rob filled the tent with balloons,

We have to find which measurement BEST describes the amount of balloons in the tent?

As tent if filled with balloons the amount of balloons equals with volume of tent.

Volume of tent is given as:

$\text { volume of tent }=a^{2}+2 a \sqrt{\frac{a^{2}}{4}+h^{2}}$

where "a" is side length and "h" is height.

$\begin{array}{l}{\text { Then, Volume }=6.3^{2}+2(6.3) \sqrt{\frac{6.3^{2}}{4}+6^{2}}} \\\\ {=39.69+12.6 \sqrt{9.925+36}} \\\\ {=39.69+12.6 \times 6.7766=39.69+85.385} \\\\ {=125.075 \mathrm{ft}^{3}}\end{array}$

Hence, from the given options, approximately, option D is correct.

6 0

3 years ago

Read 2 more answers

Find the discriminant and the number of real roots for this equation.

GuDViN [60]

Answer:

Step-by-step explanation:

To find the discriminant you do b^2-4(a)(c), which in this case gives you -28. If your discriminant is less than zero, you will have no real roots.

8 0

3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

3 years ago

Alexander was putting on a theatre show to debut his play. He is charging $2 for child

slamgirl [31]

Number of adult tickets sold is 31 and number of child tickers sold is 19

Step-by-step explanation:

Step 1: Find the equation to find the number of child and adult tickets sold.

Let the number of adult tickets sold be x, then the number of child tickets will be 50 - x. (Since there is a total of 50 seats)

Charge for a child ticket = $2

Charge for adult ticket = $5

Total amount received from show = $193

Equation ⇒ 2(50 - x) + 5x = 193

Step 2: Solve the equation to find x.

⇒ 100 - 2x + 5x = 193

⇒ 3x = 93

∴ x = 93/3 = 31

⇒ 50 - x = 50 - 31 = 19

5 0

3 years ago