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3 years ago

show that if k is a positive constant, then the area between the x-axis and one arch of the curve y=sin kx is 2/k

1 answer:

8 0

Sin(kx)=0=>x=0,π,...
∫0πsin(kx)dx=−1/kcos(kx)|π0=−1/k(cos(kπ)−cos(k∗0))
if k is odd then,=−1/k(−1−1)=−1/k(−2)=2/k
if k is even then,=−1/k(1−1)=0

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A triangle has one leg that measures 12 cm and a hypotenuse 24 cm. Find the length of the other leg to the nearest tenth of cent

Minchanka [31]

9514 1404 393

Answer:

20.8 cm

Step-by-step explanation:

The term "hypotenuse" suggests this triangle is a right triangle. Then the other leg can be found using the Pythagorean theorem:

x^2 +12^2 = 24^2

x^2 = 576 -144 = 432

x = √432 ≈ 20.8 . . . cm

The other leg is about 20.8 cm.

_____

<em>Additional comment</em>

A right triangle with a short leg that is 1/2 the length of the hypotenuse is the "special" 30°-60°-90° right triangle. The longer leg is √3 times the length of the short leg: 12√3 ≈ 20.8 cm.

7 0

2 years ago

Explain how the Quotient of Powers was used to simplify this expression. = 22 By finding the quotient of the bases to be , and c

9966 [12]

the correct question in the attached figure

(2^5)/8=2^2

we have

8----------- >2^3

(2^5)/8----------- > (2^5)/(2^3)=2^(5-3)=2^2

therefore

2^2=2^2-- ------> is ok

the answer is by simplifying 8 to 2^3 to make both powers base two and subtracting the exponents

5 0

2 years ago

ACTIVITY 2 (19) Mr Duma recently inherited a rectangular plot, part of the estate left by his late father. The plot with the fol

seraphim [82]

The sum of the lengths of three sides of the rectangle, gives the length

of the fencing, while one third of the rectangle area is for the pavement.

Responses:

Project A: The formula for the length of the fencing is, L = 4·x - 1

Project B: Length of the fancy wall = 2·x + 1

$Project \ C:Area \ of \ the \ paving = \underline{ \dfrac{2\cdot x^2 }{3} - \dfrac{x }{3} - \dfrac{1}{3}}$

<h3>Which method can be used to find the length and area of paving from the given equations?</h3>

Project A: Let QP and SR represent the longest sides of the rectangle, we have;

PQ = SR = 2·x + 1

Given parameters are;

Length of the rectangular plot = 2·x + 1

Width of the rectangular plot = x - 1

Vertices of the rectangular plot are; QPSR

Project A: Let QP and SR represent the longest sides of the rectangle, we have;

PQ = SR = 2·x + 1

Which gives;

SP = QR = x - 1

The length of the fencing, L = SP + PQ + QR = x - 1 + 2·x + 1 + x - 1 = 4·x - 1

The formula for the length of the fencing is, L =<u> 4·x - 1</u>

Project B: The front side is SR

Therefore;

Length of the fancy wall = <u>2·x + 1</u>

Project C:

Area, A = Length × Width

Area of the plot, A = (2·x + 1) × (x - 1) = 2·x² - x - 1

$Area \ of \ the \ paving = \dfrac{1}{3} \times \left(2 \cdot x^2 - x - 1\right) = \underline{ \dfrac{2\cdot x^2 }{3} - \dfrac{x }{3} - \dfrac{1}{3}}$