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3 years ago

Is (2, 5) a solution to this system of equations? 6x + y = 17 3x + 14y = 16 yes or no

Mathematics

1 answer:

Anarel [89]3 years ago

5 0

Answer:

Putting the value of x= 2 in the first equation,

6(2)+y= 17

y= 17-12 = 5

This value is same as the value for y given in the question.

Therefore, it satisfies the equation.

Again, putting the value of x= 2 in the second equation,

3(2)+14y= 16

14y= 16-6 = 10

y= 10/14 = 5/7

It doesn't satisfies the 2nd equation

Hence, (2,5) is not the solution to this system.

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assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28

melomori [17]

Given:

x and y are both differentiable functions of t.

$4x^2+3y^3=28$

$x=-1\text{ and }\dfrac{dy}{dt}=8$

To find:

The value of $\dfrac{dx}{dt}$ .

Solution:

We have,

$4x^2+3y^3=28$ ...(i)

At x=-1,

$4(-1)^2+3y^3=28$

$4+3y^3=28$

$3y^3=28-4$

$3y^3=24$

Divide both sides by 3.

$y^3=8$

Taking cube root on both sides.

$y=2$

So, y=2 at x=-1.

Differentiate (i) with respect to t.

$4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0$

Putting x=-1, y=2 and $\dfrac{dy}{dt}=8$ , we get

$4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0$

$-8\dfrac{dx}{dt}+9(4)(8)=0$

$-8(\dfrac{dx}{dt}-9(4))=0$

Divide both sides by -8.

$\dfrac{dx}{dt}-36=0$

$\dfrac{dx}{dt}=36$

Therefore, the value of $4x^2+3y^3=28$ is 36.

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2 years ago

Enter the numerical coordinates of the vertices of quadrilateral ABCD in the table. (Point A has been done for you.) Then predic

alexdok [17]

Answer:

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Step-by-step explanation:

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3 years ago

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andrey2020 [161]

Answer:

in the form 81 and 324

Step-by-step explanation:

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x=27

=> 3*27 and 12*27

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3 years ago

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kolezko [41]

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