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kari74 [83]
3 years ago
15

Help me with this!! Will mark Brainliest ❤️

Mathematics
1 answer:
Andrei [34K]3 years ago
5 0

Answer:

Yes

Step-by-step explanation:

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A roller coaster starts with the cars being pulled up a ramp. The mass of the cars is estimated by the function m(p) = 175p + 1,
KiRa [710]

Answer:

E(p)=51,450p+346,920

Step-by-step explanation:

The function

m(p) = 175p + 1,180

gives the mass of the car as a function of the number of passengers in it.

And the function

E(m) = 9.8m*h

gives the potential energy of the car as a function of the car's mass.

Now if the height of the ramp is 30 meters, we have

E(m) =9.8m*(30)

E(m) =294m

And to find the potential energy as a function of the number of passengers, we just substitute m(p) into E(m) to get:

E(m(p))=294(175p+1180)

\boxed{ E(p)=51,450p+346,920}

which gives the potential energy as a function of the number of passengers.

7 0
4 years ago
Es urgente necesito ayuda​
zalisa [80]
ANSWER

a)56
b)134
c)27
d)141
5 0
3 years ago
The function of (x) varies directly with x^2, and f(x)=96 when x=4 what is the value of f(2)
makkiz [27]
\bf \qquad \qquad \textit{direct proportional variation}
\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad  y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------

\bf \textit{f(x) varies directly with }x^2\qquad f(x)=kx^2
\\\\\\
\textit{we also know that }
\begin{cases}
f(x)=96\\
x=4
\end{cases}\implies 96=k(4)^2\implies 96=16k
\\\\\\
\cfrac{96}{16}=k\implies 6=k\qquad therefore\qquad \boxed{f(x)=6x^2}
\\\\\\
\textit{now, when }\stackrel{f(2)}{x=2}\textit{ what is \underline{f(x)}?}\qquad f(2)=6(2)^2
3 0
3 years ago
7.
anzhelika [568]
The answer is D as said above
8 0
3 years ago
In most microcomputers the addresses of memory locations are specified in hexadecimal. These addresses are sequential numbers th
iren [92.7K]

Considering that the addresses of memory locations are specified in hexadecimal.

a) The number of memory locations in a memory address range ( 0000₁₆ to FFFF₁₆ )  = 65536 memory locations

b) The range of hex addresses in a microcomputer with 4096 memory locations is ;  4095

<u>applying the given data </u>:

a) first step : convert FFFF₁₆ to decimal           ( note F₁₆ = 15 decimal )

( F * 16^3 ) + ( F * 16^2 ) + ( F * 16^1 ) + ( F * 16^0 )

= ( 15 * 16^3 ) + ( 15 * 16^2 ) + ( 15 * 16^1 ) + ( 15 * 1 )

=  61440 + 3840 + 240 + 15 = 65535

∴ the memory locations from  0000₁₆ to FFFF₁₆ = from 0 to 65535 = 65536 locations

b) The range of hex addresses with a memory location of 4096

= 0000₁₆ to FFFF₁₆ =  0 to 4096

∴ the range = 4095

Hence we can conclude that the memory locations in ( a ) = 65536 while the range of hex addresses with a memory location of 4096 = 4095.

Learn more : brainly.com/question/18993173

6 0
3 years ago
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