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2 years ago

Is the following relation a function? Yes or No

Mathematics

1 answer:

Vera_Pavlovna [14]2 years ago

7 0

Where is the relation??

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Is 4/5 bigger than 0.75

hjlf

4/5 is greater than 0.75. That is because, if you turn 4/5 into a decimal, them that decimal will be equal to 80/100. 0.75 is 75/100 as a fraction. 80/100 (4/5 simplified) compared to 75/100 is greater. The answer is yes.

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3 years ago

2. Tomás compró una bicicleta en $199.900. Primero, canceló la mitad y el resto en 7 cuotas de igual valor, con un interés total

Papessa [141]

Answer:

Cada cuota tendrá un valor de $14,850.

Step-by-step explanation:

Dado que Tomás canceló la mitad del valor de la bicicleta, la cual costaba $199.900, el valor pagado al inicio fue de $99,950 (199,900 / 2).

Luego, para el valor restante, Tomás suscribió a una financiación con un interés de $4,000, elevando el monto a pagar a $103,950, pagaderos en 7 cuotas. Por lo tanto, dichas cuotas tendrán cada una un valor de $14,850 (103,950 / 7).

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2 years ago

How many meters does 20 yards equal

Yuliya22 [10]

Answer:

Step-by-step explanation:One yard equals 0.9144 meter. Therefore 20 yards = 20 * 0.9144 = 18.288 meters

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2 years ago

Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

7 0

2 years ago

Mary's dog needs medication she told the veterinarian that the dog weighs 30 pounds the medicine is measured as 16 mg for each k

shtirl [24]

Answer:

30 pounds is 13.6078 kilograms.

13 x 16 = 208.

Mary's dog should be given 208 mg of medicine.

6 0

2 years ago