If r = 20 and t = 4, find d if d = rt. d = ???

Rob wants to draw a sector with half the arc length of the sector shown. Which method will work?

Fantom [35]

Answer: Multiply the radius,r, by 2. Divide the angle measure, x, by 4.

7 0

3 years ago

A tortilla maker produces 100 tortillas in half an hour. if he works from 9 am to 5 pm with a one hour lunch and two 15-minute b

ira [324]

The answer is 1300 i think,
hope it helps!

3 0

4 years ago

Read 2 more answers

⅛ ÷ 2 Please help! I need to remember this!

Naya [18.7K]

Answer:

1/16

Step-by-step explanation:

6 0

3 years ago

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The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to

Firdavs [7]

Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 1.

This means the area is given by the integral,

$\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy$

The integral with respect to x is trivial:

$\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy$

For the remaining integral, integrate term-by-term to get

$\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}$

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line y = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0

3 years ago

Can you please answer the question?

Roman55 [17]

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that one side of the equality is equal to the expression of the other side, requiring the use of algebraic and trigonometric properties. Now we proceed to present the corresponding procedure:

$\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}$

$\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}$

$\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}$

$\tan \alpha + 1 + \cot \alpha$

$\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1$

$\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1$

$\frac{1}{\cos \alpha \cdot \sin \alpha} + 1$

$\sec \alpha \cdot \csc \alpha + 1$

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

To learn more on trigonometric expressions: brainly.com/question/10083069

#SPJ1

6 0

2 years ago