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3 years ago

Use the function, f(x)= 2x^2+3x-2 to find f(2).

Mathematics

1 answer:

Masteriza [31]3 years ago

4 0

Answer:-9

Step-by-step explanation:

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Mr Changalls garden covers 2/5 of an acre of land .He divides the land into 4 equal sections .what fraction of an acre is each s

kakasveta [241]

1/10.
2/5 is equal to 4/10
4/10 divided by .4 is .1
so its 1/10

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3 years ago

Solve −4(x + 10) − 6 = −3(x − 2). (1 point)

AURORKA [14]

Hey there!!

Given equation :

... -4 ( x + 10 ) - 6 = -3 ( x - 2 )

Using the distributive property :

... -4x - 40 - 6 = -3x + 6

Combining like terms :

... -4x - 46 = -3x + 6

Adding 46 on both sides :

... -4x = -3x + 52

Adding 3x on both sides :

... -x = 52

... x = -52

Hope helps!

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3 years ago

What is the greatest common factor of 10 and 50

brilliants [131]

The greatest number common to 50 and 10 is 10 10 is the answer hope this helps

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2 years ago

Read 2 more answers

an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a

viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is; $\mathbf{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at a time t and r = radius of the cone shaped at a time t

Then the similar triangles ΔOCD and ΔOAB is as follows:

$\dfrac{h}{r}= \dfrac{20}{8}$ ( similar triangle property)

$\dfrac{h}{r}= \dfrac{5}{2}$

$\dfrac{h}{r}= 2.5$

h = 2.5r

$r = \dfrac{h}{2.5}$

The volume of the water in the tank is represented by the equation:

$V = \dfrac{1}{3} \pi r^2 h$

$V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h$

$V = \dfrac{1}{18.75} \pi \ h^3$

The rate of change of the water depth is :

$\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

Since the water is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

$\dfrac{dv}{dt}= - 4 \ ft^3/sec$

Therefore,

$-4 = \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

the rate of change of the water at depth h = 10 ft is:

$-4 = \dfrac{ 100 \ \pi }{6.25}\ \dfrac{dh}{dt}$

$100 \pi \dfrac{dh}{dt} = -4 \times 6.25$

$100 \pi \dfrac{dh}{dt} = -25$

$\dfrac{dh}{dt} = \dfrac{-25}{100 \pi}$

Thus, the rate of change of the water depth when the water depth is 10 ft is; $\mathtt{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

4 0

3 years ago

Please help with this

julsineya [31]

Here's the work and the answer. I hope this helps.

5 0

3 years ago