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abruzzese [7]
3 years ago
6

RADIO TELESCOPES The cross section of a large radio telescope is a parabola. The equation that describes the cross section is y=

275x2−43x−323,
where y is the depth of the dish in meters at a point x meters from the center of the dish. If y=0
represents the top of the dish, what is the width of the dish? Solve by graphing.
Mathematics
1 answer:
Minchanka [31]3 years ago
3 0

Answer:

I am sorry

Step-by-step explanation:

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y is directly proportional to the positive square root of x when x = 9, y= 12 find y when x =1/2​
ehidna [41]

Y =k√x

12 = 3k

k = 4

now X = 1\2

y = 4√1/2

y = 2√2

6 0
3 years ago
Read 2 more answers
How to differentiate y=x^n using the first principle. In this question, I cannot use the rule of differentiation. I have to do t
Zarrin [17]

By first principles, the derivative is

\displaystyle\lim_{h\to0}\frac{(x+h)^n-x^n}h

Use the binomial theorem to expand the numerator:

(x+h)^n=\displaystyle\sum_{i=0}^n\binom nix^{n-i}h^i=\binom n0x^n+\binom n1x^{n-1}h+\cdots+\binom nnh^n

(x+h)^n=x^n+nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n

where

\dbinom nk=\dfrac{n!}{k!(n-k)!}

The first term is eliminated, and the limit is

\displaystyle\lim_{h\to0}\frac{nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n}h

A power of h in every term of the numerator cancels with h in the denominator:

\displaystyle\lim_{h\to0}\left(nx^{n-1}+\dfrac{n(n-1)}2x^{n-2}h+\cdots+nxh^{n-2}+h^{n-1}\right)

Finally, each term containing h approaches 0 as h\to0, and the derivative is

y=x^n\implies y'=nx^{n-1}

4 0
3 years ago
What quantum numbers specify these subshells 5s 2p 3d?
Ivan
5s subshell represent that n is equal to 5 and l is equal to 0.
2p subshell represent that n is equal to 2 and l is equal to 1.
3d subshell represent that n is equal to 3 and l is equal to 2.
The numbers gives us the value of n and the letter gives us the value of l.
s means l = 0 
p means l = 1 
d means l = 2
f means l = 3 
5 0
3 years ago
I WILL MARK YOU THE BRAINLIEST IF YOU ANSWER THIS CORRECTLY WITH WORK
saul85 [17]

Answer:

Students on Honor Roll had no advantage to get the Science class requested. It was a fair class assignment. In fact No Honor Roll students had advantage over honor Roll Students in obtaining Science Class as requested.  

Step-by-step explanation:

Here let us streamline the information we are given one by one

1.) Honor Roll

Get Requested Science Class                : 64

Did Not get  Requested Science Class : 80

Total                                                         : 144

Percentage of Honor Roll Students getting Science class as requested =

\frac{64}{144}

= 44.44%

2.) No Honor Roll

Get Requested Science Class               : 315

Did Not get Requested Science Class :  41

Total                                                         : 356

Percentage of Non Honor Roll Students getting Science class as requested = \frac{315}{356}

= 88.8%

By observing the above result , we can assume that Honor Roll students did not have any advantage over non Honor Roll students in getting Science Class as they requested.

3 0
3 years ago
A bag contains 6 RED beads, 3 BLUE beads, and 11 GREEN beads. If a single bead is picked at random, what is the probability that
Anna11 [10]

Answer:

17/20

Step-by-step explanation:

Total no. of beads = 6 + 3 + 11 = 20

We need RED or GREEN bead .

So no. of beads needed = 6 + 11 = 17

So probability of getting GREEN or RED beads = 17/20

6 0
3 years ago
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