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3 years ago

Evaluate 2^-4 pleaseee help asap

Mathematics

2 answers:

Alexxandr [17]3 years ago

5 0

Answer:

1/16

Step-by-step explanation:

2^-4

We know that a^ -b = 1/ a^b

2^-4 = 1/2^4

2^4 =16

so 1/2^4 = 1/16

vichka [17]3 years ago

3 0

Answer:

2 ^-4 = 1/16

Step-by-step explanation:

2^-4 = 1/2^4

2^4 =16

so 1/2^4 = 1/16

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The base of a parallelogram is 18 inches longer than three times the height. The area of the parallelogram is 216 square inches.

nordsb [41]

Answer:

12 inches

Step-by-step explanation:

Let b represent the base

h represents the height

area of the parallelogram = base * height = 216 square inches

From the question'

b = 18 + 3h

Slot in the value of b

216 = (18 + 3h) * h

expand

216 = 18h + 3h^2

subtract 216 from both sides

0 = 18h + 3h^2 - 216

rearrange

3h^2 + 18h - 216 = 0

divide through by 3

h^2 + 6h - 72 = 0

Now, lets solve!

h^2 + 6h - 12h - 72 = 0

h( h + 6 ) - 12(h + 6) = 0

(h - 12) (h + 6) = 0

h - 12 = 0

h = 12

and

h + 6 = 0

h = - 6

Taking the positive value of h

Hence, the height is 12 inches

Lets check

when h = 12 inches

Area of the parallelogram = 18* 12 = 216 square inches .... correct

when h = -6inches

A = 18 * -6 ≠ 216 square inches

So height is 12 inches

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3 years ago

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3 years ago

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Step-by-step explanation:

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3 years ago

the local phone company charges 71.32 keep your phone hooked up and 0.50 for everything cause you make if you make 150 calls how

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You will own the company for 146.32

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3 years ago

Find the points on x^2+3(y-2)^2 = 27 which are closest to and farthest from the origin

maria [59]

$(-4.89, 3)$ and $(4.89, 3)$ are the farthest from the origin and $(0, -1)$ is the closest from the origin.

Let be a curve of the form $x^{2}+3\cdot (y-2)^{2} = 27$ . The distance with respect to origin is found by using the following Pythagorean identity:

$r = x^{2}+y^{2}$ (1)

Where $r$ is the square distance function.

We can modify (1) as following:

$r = [27-3\cdot (y-2)^{2}]+y^{2}$

$r = 27-3\cdot (y^{2}-4\cdot y +4)+y^{2}$

$r = 27-3\cdot y^{2}+12\cdot y -12 + y^{2}$

$r = -2\cdot y^{2}+12\cdot y+27$ (1b)

Now we apply the First and second derivative tests to determine the minimum and maximum distances from the origin:

First derivative test

$-4\cdot y +12 = 0$

$y = 3$

Second derivative test

$r'' = -4$

The y-component represents a maximum.

Now we graph the function with the ressource of a graphing tool, we find the following points:

Farthest points: $(-4.89, 3)$ , $(4.89, 3)$ .

Closest points: $(0, -1)$ .

$(-4.89, 3)$ and $(4.89, 3)$ are the farthest from the origin and $(0, -1)$ is the closest from the origin.

To learn more on ellipses, we kindly invite to check this verified question: brainly.com/question/19507943

6 0

3 years ago