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katrin [286]
3 years ago
10

F(x) = 3x^2+5x-14 Find f(-9)

Physics
1 answer:
Llana [10]3 years ago
8 0

Answer:

f(-9) = 184

Explanation:

f(x)=3x²+5x-14

f(-9)= 3(-9)² +5(-9)-14 Order of Operations : Exponents

     =  3(81)+5(-9)-14

     = 243+5(-9)-14

     =  243-45-14

     = 198-14

f(-9)= 184

Hope this helps :)

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The diameter of a hydrogen atom is 0.000000000106 m. How can this
Andreas93 [3]

Answer:

The diameter of a hydrogen atom based on scientific notation is 1.06 x 10^-10 m

8 0
3 years ago
A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow
Ksivusya [100]

Answer:

The answer is 1.87nm/s.

Explanation:

The 110g/hr  water loss must be replaced by 110g/hr of sap. 110g of sap corresponds to a volume of  

110g \div \dfrac{1040*10^3g}{1*10^6cm^3}  = 106cm^3

thus rate of sap replacement is

106cm^3/hr = 106*10^{-6}m^3/3600s  = 2.94*10^{-8}m^3/s

The volume of sap in the vessel of length x is

V = Ax,

where A is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

V = 2000Ax

taking the derivative of both sides we get:

\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}

on the left-hand-side \dfrac{dx}{dt} is the velocity v of the sap, and on right-hand-side \dfrac{dV}{dt}  = 2.94*10^{-8}m^3/s; therefore,

2.94*10^{-8}m^3/s=2000Av

and since the cross-sectional area is

A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2;

therefore,

2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v

solving for v we get:

v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}

\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}

which is the upward speed of the sap in each vessel.

8 0
3 years ago
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Luba_88 [7]

I could make a poem for you if you actually gave the words...... what 10 words do i need to incorporate???☹︎

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3 years ago
I need help ASAP plzzzz
Fiesta28 [93]

Answer:

a) 1.75s b) 17.2 m/s (down)

Explanation:

d1= 15m d2= 0m (because it hits ground)

a= -9.81 m/s^2 t=???

Equation

the triangle means change in so d2-d1

Δd= v1 * t + 1/2 * a * t^2

0m-15m= v1*t + 1/2 a t^2

-15 m= 0m/s*t (goes away) + 1/2* a *t^2

-15mx2= t^2

-15mx2/a= t^2

Square root (-30/-9.81m/s^2)

t=1.75 s

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Im going to use v2= v1 + a*t

v2= 0m/s + -9.81 x 1.75s

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3 years ago
4. The value of the before and after-collision momentum of two colliding objects is shown in the
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