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3 years ago

13

a rectangular red sticker has a perimeter of 20 millimeters . it's area is 21 square millimeters . what are the dimensions of th

e sticker?

1 answer:

Delvig [45]3 years ago

6 0

Answer:

the stick is a 7 by 3

Step-by-step explanation:

7+3+7+3 adds to 20 which would be the perimeter and the area is just length times width so 7 times 3.

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What is the maximum number of possible extreme values for the function f(x)=x^4+x^3-7x^2-x+6

notka56 [123]

A 4th degree polynomial will have at most 3 extreme values. Since the degree is even, there will be one global extreme, with possible multiplicity. The remainder, if any, will be local extremes that may be coincident with each other and/or the global extreme.

(The number of extremes corresponds to the degree of the derivative, which is 1 less than the degree of the polynomial.)

3 0

3 years ago

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Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A<br>What is the value of the A^2?<br><br>

Alla [95]

$\large \mathbb{PROBLEM:}$

$\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}$

$\large \mathbb{SOLUTION:}$

$\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}$

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4 0

2 years ago

What is the area of the shaded region below?<br> 7 m<br> 30 m<br> 7 m<br> 21 m

Rama09 [41]

Answer:

322.28

A

Step-by-step explanation:

Area of the rectangle

L = 30

W = 21

Area = 30 * 21

Area = 630

Area of the circles

Radius = 7 m

Area = pi r^2

Area = 3.14 * 7^2

Area = 153.86 m^2

Area of both circles = 307.72

Area of the Shaded Region

Area of shaded region = area of the rectangle - area of 2 circles.

Area of Shaded Region = 630 - 307.72

Area of Shaded Region = 322.28

5 0

2 years ago

A 12 ft ladder is leaning against a building. The ladder makes a 45 degree angle with the building how far up is the building do

Ghella [55]

Answer:

The ladder reach a height 8.485 ft.

Step-by-step explanation:

The ladder with the building made right angle triangle

The length of the ladder is the hypotenuse 12 ft.

The angle between the ladder and the building is 45°

The height of the ladder reach is the vertical side of the Δ

∴ h = 12 cos 45° = 6√2 = 8.485 ft.

4 0

3 years ago

Find the perimeter and area of a rectangle of cardboard measuring 14.1 cm by 6.5 cm

allsm [11]

Answer:

perimeters =41.2

area=91.65

Step-by-step explanation:

rectangle Perimeter(<em>p</em>) Area (<em>A</em>)

L=length 2 (L+W) L×W

W=width

8 0

3 years ago

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