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3 years ago

Given the relation {(1, 4), (3, 6), (5, 4), (1, 8)},

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

5 0

Answer:

The domain of relation is {1,3,5}

It is not a function.

Step-by-step explanation:

Given the relation {(1, 4), (3, 6), (5, 4), (1, 8)}

We need to find the domain and determine if the given relation is function or not.

The domain of relation is {1,3,5}

But the given relation is not function because we have x=1 twice in the relation and the for the relation to be function the values of domain can't be repeated.

So, It is not a function.

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Mrs.Kunkleman asks her students to write the Pythagorean theorem in their own words

masha68 [24]

Step-by-step explanation:

The Pythagorean theorem can be understood as a mathematical relationship between the sides of a right triangle that helps to understand geometric problems in real situations, such as finding measurements, calculating areas, etc.

The theorem says that the square of the hypotenuse is equal to the sum of the squares on the other sides.

In a right triangle the hypotenuse is the longest side of the triangle, on the side opposite to its longest angle, and the other two sides will be the sides. So the Pythagorean theorem formula is:

a² = b² + c²

where a represents the hypotenuse and b and c the other sides.

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3 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6) Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

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2 years ago

6n+n+14=0 i need the math

Fofino [41]

Let's solve your equation step-by-step.6n+n+14=0

Step 1: Simplify both sides of the equation.6n+n+14=0Simplify: (Show steps)7n+14=0

Step 2: Subtract 14 from both sides.7n+14−14=0−147n=−14

Step 3: Divide both sides by 7.7n7=−147n=−2

Answer:n=−2

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3 years ago

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Fynjy0 [20]

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Answer is B.

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3 years ago

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The composite figure is made up of three simpler shapes. What is the area of the figure?

OlgaM077 [116]

Answer:

C) 55 cm^2

Step-by-step explanation:

(12x3)+(8x2)+(1/2)(3)(2) = 55

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2 years ago