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Cerrena [4.2K]
3 years ago
9

Written as a simplified polynomial in standard form, what is the result when

Mathematics
1 answer:
arlik [135]3 years ago
5 0

8x3 + 10x2 – 8x + 1

: Hope this helps :)
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Cameron is designing a calendar as a fund raising project for math class. The cost of printing is $500, plus $2.50 per calendar.
belka [17]
Y = 2.5x - 500

I simplified it a bit, you could write it as y = 5x - 2.5x - 500
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This problem focuses on using Polymath, an ordinary differential equation (ODE) solver, and also a non-linear equation (NLE) sol
dezoksy [38]

Answer:

here

Step-by-step explanation:

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3 years ago
What is 1 3/5 times 3/4?
Bingel [31]

Answer:

1.2

Step-by-step explanation:

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4 years ago
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Please resolve this problem for me! I've been having trouble with this for a while.
Bad White [126]

Answer:

a) 41

b) -31

c) 25/3

Step-by-step explanation:

The instructions explain what a function is pretty clearly.

Basically, whatever is in the parentheses next to the f, you plug in to the expression that is the function.

ex:     f(x) = x + 4

        f(5) = 5 + 4 = 9

Okay, onto the problems:

a) f(x) = 3x - 1

First, plug in 14, since that is given:

f(14) = 3(14) - 1

and solve.

f(14) = 42 - 1

f(14) = 41

So, 41 is the answer to a.

b) Do the same thing you did in problem a. Plug in the number inside the parentheses and solve.

f(-10) = 3(-10) - 1

f(-10) = -30 - 1

f(-10) = -31

-31 is the answer to b.

c) For this problem, you have to do a bit of algebra.

First, take the base function f(x):

f(x) = 3x - 1

Then, set f(x) = 24:

24 = 3x - 1

and solve by isolating x.

25 = 3x

\dfrac{25}{3} = x

x = \dfrac{25}{3}

25/3 is the answer to c.

8 0
2 years ago
What is the maximum number of consecutive odd positive integers that can be added together before the sum exceeds $401$
salantis [7]

The maximum number of integers is 27 that can be added together before the summation of the A.P. Series exceeds 401.

According to the statement

We have a given that the maximum sum of the positive integers is 400.

And we have to find the value of n which is a maximum number of integers by which the value of sum become 400.

So, to find the value of the n we use the

A.P. Series'Summation formula

According to this,

S = n (n+1)/2

Here the value of s is 401

Then

S = n (n+1)/2

401 = n (n+1)/2

401*2 = n (n+1)

802 =n (n+1)

n (n+1) = 802

n^2 + n -802 =0

By the use of the Discriminant formula the

value of n becomes n = -28 and n = 27.

The negative value of n is neglected.

Therefore the value of n is 27.

So, The maximum number of integers is 27 that can be added together before the summation of the A.P. Series exceeds 401.

Learn more about maximum number of integers here brainly.com/question/24295771

#SPJ4

7 0
2 years ago
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