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3241004551 [841]
4 years ago
8

What is 3x^3-8x^3-5x+6 subtracted from x^4+3x^2+5x*6

Mathematics
1 answer:
Nikitich [7]4 years ago
3 0

Answer: x^4 - 5x^3 + 3x^2 + 25x + 6

Step-by-step explanation: Simplify the expression.

Hope this helps you out! ☺

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A soccer goalie saved 76% of 225 goal attempts last season. How many saves did the goalie successfully make?
tamaranim1 [39]
The player made 171 goals

7 0
3 years ago
Find the equation of the line passing through (4,3) and parallel to the line whose equation is y+2x=m
Reptile [31]

Answer:

y = - 2x + 11

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Rearrange y + 2x = m into this form by subtracting 2x from both sides

y = - 2x + m ← in slope- intercept form

with slope m = - 2

Parallel lines have equal slope, thus

y = - 2x + c ← is the partial equation

To find c substitute (4, 3) into the partial equation

3 = - 8 + c ⇒ c = 3 + 8 = 11

y = - 2x + 11 ← equation of line

5 0
3 years ago
Read 2 more answers
In each situation, write a recurrence relation, including base case(s), for the given function. briefly explain in words why thi
vova2212 [387]
A thumb-wrestling match requires two thumbs, so we can either suppose that we need at least two people (n\ge2), or allow one to thumb-wrestle one's self. In either case, we'd have t(1)=t(2)=1, so let's just say we need a minimum of two players.

If we add one more person to the set of players, then the first two people would need to play 1 additional match each. So t(3)=t(2)+2.

If we add one more person, then the first three people would again each have to play 1 more match with the new person. So t(4)=t(3)+3.

And so on, so that in general, the number of games needed for everyone to play exactly one match with everyone else is given recursively by

\begin{cases}t(2)=1\\t(n+1)=t(n)+n&\text{for }n\ge2\end{cases}
6 0
3 years ago
I need help please!!
Jobisdone [24]
A
B
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F
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(It’s basically just a game of making pairs)
4 0
3 years ago
I need to find the sum of this equation
Tresset [83]

Answer:

\large\boxed{\dfrac{3s^2+19s-4}{(s-1)^2(s+5)}=\dfrac{3s^2+19s-4}{s^3+3s^2-9s+5}}

Step-by-step explanation:

s^2-2s+1=s^2-s-s+1=s(s-1)-1(s-1)=(s-1)(s-1)=(s-1)^2\\\\s^2+4s-5=s^2+5s-s-5=s(s+5)-1(s+5)=(s+5)(s-1)\\\\\text{Therefore}\ LCD\ \text{is}\ (s-1)^2(s+5).

\dfrac{3s}{s^2-2s+1}+\dfrac{4}{s^2+4s-5}=\dfrac{3s(s+5)}{(s-1)^2(s+5)}+\dfrac{4(s-1)}{(s-1)^2(s+5)}\\\\\text{use the distributive property}\\\\=\dfrac{(3s)(s)+(3s)(5)+(4)(s)+(4)(-1)}{(s-1)^2+(s+5)}=\dfrac{3s^2+15s+4s-4}{(s-1)^2(s+5)}\\\\=\dfrac{3s^2+19s-4}{(s-1)^2(s+5)}

4 0
3 years ago
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