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s2008m [1.1K]
3 years ago
7

Find domain and range of 2x+3y=4

Mathematics
1 answer:
jeka943 years ago
7 0

Answer:

The domain is (-infinity, infinity)

The range is (-infinity, infinity)

Step-by-step explanation:

In order to turn this equation into a linear function, you must solve for y, to get y=. First you subtract 2x from both sides. You will then have 3y=4-2x. Next, you divide by 3 on both sides to isolate the "y".  Then you'll get y=-2/3x+4/3. The domain and range of this function once it is graphed is (-infinity, infinity).

Hope I helped :)

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17 please help me......................
Vaselesa [24]
I think its (6.5+3.25)÷0.2=48.75

dont quote me on that, I tried and I'm not the best at math lol.
(6.5 + 3.25) \div 0.2
8 0
3 years ago
Simply 11a+7+10a+3+12a+9 by combining like terms​
patriot [66]

Answer: 33a+19

Step-by-step explanation:

11a+7+10a+3+12a+9

Collect like terms.

(11a+10a+12a)+(7+3+9)

Simplify.

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6 0
3 years ago
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Haya la integral de xarcsenx/((1-x^2)^1/2). Utiliza la sustitución de t= arcsenx
ikadub [295]

Answer:

<em>Explicación más abajo</em>

Step-by-step explanation:

Integración Indefinida

La integral

I=\int \dfrac{x .arcsen\left(x\right)}{\sqrt{1-x^2}}

Se resuelve con el cambio de variables:

t=arcsen(x)

Una vez hechos los cambios, la integral se resuelve en función de t:

I=sen(t)-t.cos(t)+C

Hay que devolver los cambios para mostrarla en función de x.

El cambio de variables también se puede escribir:

x=sen(t)

Recordando que

cos(t)=\sqrt{1-sen^2(t)}

Entonces:

cos(t)=\sqrt{1-x^2}

Devolviendo los cambios:

I=sen(t)-t.cos(t)+C=x-arcsen(x)\sqrt{1-x^2}+C

Es la respuesta correcta

7 0
3 years ago
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Zarrin [17]

Answer:

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Step-by-step explanation:

The equation of a circle is given by

(x-h)^2+(y-k)^2 = r^2

Where (h,k) is the center and r is the radius

(x-0)^2+(y-0)^2 = 3^2

The center is (0,0) and the radius is 3

3 0
3 years ago
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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br>Which expression is equal to 5/13√ ?
Anvisha [2.4K]

\dfrac{5}{\sqrt{13}}\cdot\dfrac{\sqrt{13}}{\sqrt{13}}=\dfrac{5\sqrt{13}}{13}

\text{Used:}\ \sqrt{a}\cdot\sqrt{a}=a

7 0
3 years ago
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