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3 years ago

12

The function f(x)=(2x-3)(x+k).Given that f(0)=15. Calculate

1 answer:

Valentin [98]3 years ago

4 0

Answer:

(a-b)

Step-by-step explanation:

a minus b whole square minus C square

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A radio station is going to construct a 5-foot tower on top of a building. the tower will be supported by three cables, each att

KonstantinChe [14]

You need to use Pythagoras’ Theorem.
A² + B² = c² and c is the hypotenuse of the right angled triangle.

The height of the tower (5 feet) is A and the distance from the end of the cable and the base of the tower (12 feet) is B. The length of ONE cable is c. So:

A² + B² = c²
5² + 12² = c²
25 + 144 = c²
169 = c²
13 = c. This is the length of one cable.

3 x 13 = 39 therefore the total length of the cables is 39 feet.

25

8 0

3 years ago

(How do I solve <br> 1/3)^4

Serga [27]

Answer:

............ .. ..sjsjjs

7 0

3 years ago

Read 2 more answers

The width and length of a rectangle (in feet)are consecutive odd integers. If the length is increased by 5 feet, the area of the

hodyreva [135]

The actual (late) answer is 35ft^2

8 0

3 years ago

22/100 in simplest form

lubasha [3.4K]

22/100:

You can actually divided the numerator and denominator by a number that can go into both 22 and 100 but also not get a remainder.

So 22/2= 11

And 100/2=50

Combined you get 11/50

Hope it helps!!!

5 0

3 years ago

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The polar curve $r = 1 + \cos \theta$ is rotated once around the point with polar coordinates $(2,0).$ What is the area of the r

mash [69]

Answer:

$Area = -2.3147$

Step-by-step explanation:

Given

$r = 1 + \cos \theta$

Required

Determine the area with coordinates $(2,0)$

The area is represented as:

$Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta$

Where

$r = 1 + \cos \theta$

and

$(a,b) = (2,0)$

Substitute values for r, a and b in

$Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta$

$Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)^2} \, d\theta$

Expand

$Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)(1 + cos\theta)} \, d\theta$

$Area = \frac{1}{2}\int\limits^0_2 {(1 + 2cos\theta+cos^2\theta} )\, d\theta$

By integratin the above, we get:

$Area = \frac{1}{2}*\frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{2}[0,2]$

$Area = \frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{4}[0,2]$

Substitute 0 and 2 for $\theta$ one after the other

$Area = \frac{(cos(0) + 4)sin(0) + 3*0}{4} - \frac{(cos(2) + 4)sin(2) + 3*2}{4}$

$Area = \frac{(cos(0) + 4)sin(0)}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = \frac{(1 + 4)*0}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = \frac{-sin(2)(cos(2) + 4) - 6}{4}$

Get sin(2) and cos(2) in radians

$Area = \frac{-0.9093 * (-0.4161 + 4) - 6}{4}$

$Area = \frac{-9.2588}{4}$

$Area = -2.3147$

3 0

2 years ago