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denis23 [38]
3 years ago
13

What is the constant of the polynomial 2x3 - 8x2 + 3x - 7

Mathematics
1 answer:
Firlakuza [10]3 years ago
8 0

Answer:

-7

Step-by-step explanation:

A constant number is a number that contains no variables like x and y. The only constant in that problem is -7.

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If a sphere has a volume of 9198.1 in, what is the radius of the sphere?
KonstantinChe [14]

Answer:

13

Step-by-step explanation:

Volume to find radius: (3\frac{V}{4\pi})^1^/^3

Substitute: (3\frac{9198.1}{4\pi})^1^/^3

Solve: (3(732.3))^1^/^3

2196.9^1^/^3

13≈r

5 0
2 years ago
Which real-world scenario can be described by the algebraic expression StartFraction p over 8 EndFraction?
Basile [38]

Answer:

the answer is D

Step-by-step explanation:

I just took a test

8 0
3 years ago
Read 2 more answers
Pwease help me anyone. thanks :-) :-) :-) :-) :-) :-) :-) :-)!!!!
adell [148]
A)
x+3

b)
x(x+3)=340\\
x^2+3x-340=0

c)
x(x+3)=340\\ x^2+3x-340=0\\
\Delta=3^2-4\cdot1\cdot(-340)=9+1360=1369\\\sqrt{\Delta}=37\\
x_1=\dfrac{-3-37}{2}=-20\\
x_2=\dfrac{-3+37}{2}=17

It's 17.
3 0
3 years ago
What is the measure of the inscribed angle ABC if the measure of the arc, which this angle intercepts is:
Margaret [11]
I believe the answer should be A
5 0
3 years ago
7. A large population of ALOHA users manages to generate 60 requests/s, including originals and retransmissions. Time is slotted
Debora [2.8K]

Answer:

P(success at first attempt) = 0.1353

Step-by-step explanation:

This question follows poisson distribution. Thus, the formula is;

P(k) = (e^(-G) × (G)k)/k!

where;

G is number of frames generated in one frame transmission time(or frame slot time)

Let's find G.

To do this, we need to find number of frames generated in 1 slot time which is given as 50 ms.

Now, in 1000 ms, the number of frames generated = 50

Thus; number of frames generated in 50 ms is;

G = (50/1000) × 50

G = 2.5

To find the chance of success on the first attempt will be given by;

P(success at first attempt) = P(0) = e^(-G) = e^(-2) = 0.1353

8 0
2 years ago
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