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2 years ago

Whats the volume 14cm 10cm __cm3

Mathematics

1 answer:

Klio2033 [76]2 years ago

5 0

Answer:

The volume of the cuboid = 4 × 2 × 3 = 24 cm3. For any ... 3 A brick is 20 cm long, 12 cm wide and 10 cm high. What is its volume? 20 cm. 12 cm. 10 cm. 5 m. 3 m

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A conical circus tent has a 20 ft central pole that supports it. the slant height of the tent is 26 ft long. explain how to find

Degger [83]

The angle the pole makes with the tent is 39.7°.

<h3>What is an angle?</h3>

An angle is a figure in Euclidean geometry created by two rays, called the sides of the angle, that share a common termination, called the vertex of the angle.
Angles formed by two rays are located in the plane containing the rays.

Trigonometric ratio: the trigonometric ratio is used to demonstrate the relationship between the sides and angles of a right-angled triangle.

Let θ represent the angle the pole makes with the tent.

Using trigonometric ratios:

cos(θ) = 20/26
θ = 39.7°

Therefore, the angle the pole makes with the tent is 39.7°.

Know more about trigonometric ratios here:

brainly.com/question/24349828

#SPJ4

4 0

1 year ago

A) Findi

Romashka [77]

Answer:

Part A)

$\displaystyle \frac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}$

Part B)

$\displaystyle y=-\frac{5}{8}x+\frac{9}{4}$

Step-by-step explanation:

We have the equation:

$\displaystyle x^2y+y^2x=6$

Part A)

We want to find the derivative of our function, dy/dx.

So, we will take the derivative of both sides with respect to <em>x:</em>

<em /> $\displaystyle \frac{d}{dx}\Big[x^2y+y^2x\Big]=\frac{d}{dx}\big[6\big]$ <em />

The derivative of a constant is 0. We can expand the left:

$\displaystyle \frac{d}{dx}\Big[x^2y\Big]+\frac{d}{dx}\Big[y^2x\Big]=0$

Differentiate using the product rule:

$\displaystyle \Big(\frac{d}{dx}\big[x^2\big]y+x^2\frac{d}{dx}\big[y\big]\Big)+\Big(\frac{d}{dx}\big[y^2\big]x+y^2\frac{d}{dx}\big[x\big]\Big)=0$

Implicitly differentiate:

$\displaystyle (2xy+x^2\frac{dy}{dx})+(2y\frac{dy}{dx}x+y^2)=0$

Rearrange:

$\displaystyle \Big(x^2\frac{dy}{dx}+2xy\frac{dy}{dx}\Big)+(2xy+y^2)=0$

Isolate the dy/dx:

$\displaystyle \frac{dy}{dx}(x^2+2xy)=-(2xy+y^2)$

Hence, our derivative is:

$\displaystyle \frac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}$

Part B)

We want to find the equation of the tangent line at (2, 1).

So, let's find the slope of the tangent line using the derivative. Substitute:

$\displaystyle \frac{dy}{dx}_{(2,1)}=-\frac{2(2)(1)+(1)^2}{(2)^2+2(2)(1)}$

Evaluate:

$\displaystyle \frac{dy}{dx}_{(2,1)}=-\frac{4+1}{4+4}=-\frac{5}{8}$

Then by the point-slope form:

$y-y_1=m(x-x_1)$

Yields:

$\displaystyle y-1=-\frac{5}{8}(x-2)$

Distribute:

$\displaystyle y-1=-\frac{5}{8}x+\frac{5}{4}$

Hence, our equation is:

$\displaystyle y=-\frac{5}{8}x+\frac{9}{4}$

5 0

1 year ago

Consider the integral 8 (x2+1) dx 0 (a) Estimate the area under the curve using a left-hand sum with n = 4. Is this sum an overe

Elina [12.6K]

Answer:

(a) 120 square units, underestimate

(b) 248 square units, overestimate

Step-by-step explanation:

(a) <u>left sum</u>

The left sum is the sum of the areas of the rectangles whose width is the total interval width (8-0) divided by the number of divisions (n=4). The height of each rectangle is the function value at its left edge.

We can compute the sum by adding the function values and multiplying that total by the width of the rectangles:

left sum = (1 + 5 + 17 + 37)×2 = 60×2 = 120 . . . square units

The curve is increasing throughout the interval of interest, so the left sum underestimates the area under the curve.

(b) <u>right sum</u>

The rectangles whose area is the right sum are shown in the attachment, along with the table of function values. The right sum is computed the same way as the left sum, but using the function value on the right side of each subinterval.

right sum = (5 + 17 + 37 + 65)×2 = 124×2 = 248 . . . square units

The curve is increasing throughout the interval of interest, so the right sum overestimates the area under the curve.

_____

The actual area under the curve on the interval [0, 8] is 178 2/3, just slightly less than the average of the left- and right- sums.