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3 years ago

13

Please help me with this.

1 answer:

pshichka [43]3 years ago

6 0

Given:

The quadratic equation is:

$3x^2-5x-2=0$

To find:

Point that is the solution of given equation.

Solution:

We have,

$3x^2-5x-2=0$

The quadratic function of the above equation is:

$y=3x^2-5x-2$

Putting $x=1$ , we get

$y=3(1)^2-5(1)-2$

$y=3-5-2$

$y=-4$

The function passes through the point (1,-4). It means (1,-4) is the solution of given equation but (1,5) is not a solution of given equation.

Therefore, the correct option is A.

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The surface area of a cylinder is increasing at a rate of 9 pi square meters per hour. The height of the cylinder is fixed at 3

Alekssandra [29.7K]

Answer:

$9\pi \text{ cubic meters per hour}$

Step-by-step explanation:

Since, the surface area of a cylinder,

$A= 2\pi r^2 + 2\pi rh$ ................(1)

Where,

r = radius,

h = height,

If $A= 36\pi\text{ square meters}, h = 3\text{ meters}$

$36\pi = 2\pi r^2 + 2\pi r(3)$

$18 = r^2 + 3r$

$\implies r^2 + 3r - 18=0$

$r^2 + 6r - 3r - 18 = 0$ ( by middle term splitting )

$r(r+6)-3(r+6)=0$

$(r-3)(r+6)=0$

By zero product property,

r = 3 or r = - 6 ( not possible )

Thus, radius, r = 3 meters,

Now, differentiating equation (1) with respect to t ( time ),

$\frac{dA}{dt}= 4\pi r\frac{dr}{dt} +2\pi(r\frac{dh}{dt} + h\frac{dr}{dt})$

∵ h = constant, ⇒ dh/dt = 0,

$\frac{dA}{dt} = 4\pi r \frac{dr}{dt} +2\pi h \frac{dr}{dt}$

We have, $\frac{dA}{dt}=9\pi\text{ square meters per hour}, r = h = 3\text{ meters}$

$9\pi = 4\pi (3) \frac{dr}{dt}+2\pi (3)\frac{dr}{dt}$

$9\pi = (12\pi + 6\pi )\frac{dr}{dt}$

$9\pi = 18\pi \frac{dr}{dt}$

$\implies \frac{dr}{dt} =\frac{1}{2}\text{ meter per hour}$

Now,

Volume of a cylinder,

$V=\pi r^2 h$

Differentiating w. r. t. t,

$\frac{dV}{dt}=\pi ( r^2 \frac{dh}{dt}+h(2r)\frac{dr}{dt})=\pi ((3)(6) (\frac{1}{2})) = 9\pi \text{ cubic meters per hour}$

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3 years ago

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erastova [34]

Answer:

the first one is 37

the second one is 36

the third one is 27

Step-by-step explanation:

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