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maxonik [38]
3 years ago
7

write the 3 terms of ( 2a+ax)^5 given the first terms in the expansion (b +2x) (2+ ax)^5 are 96 - 176x+cx^2. find the values of

a,b,c
Mathematics
1 answer:
Sloan [31]3 years ago
7 0

Answer:

a^5x^5, 10a^5x^4, 40a^5x^3

Step-by-step explanation:

Use pascal's triangle for the first one

(2+x)^5 * a^5

= x^5a^5 + 5*2^1*x^4*a^5 + 10*2^2*x^3*a^5 ...

= a^5x^5 + 10a^5x^4+ 40a^5x^3 ...

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Which number is represented by point E on the graph?
zavuch27 [327]

Answer:

It'll be -0.25

Step-by-step explanation:

Notice how it is to the left of zero, anything to the left of zero is negative. With that in mind, we can eliminate A. The half mark between 0 and -1 represents 0.50, and since it goes by quarters, the E is at the first quarter, which is 0.25. Hopefully this helps!

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In the graph below, Point A represents Owen's house, Point B represents David's house and Point C represents the school. Who liv
ycow [4]

Answer:

• David

,

• 4 miles

Explanation:

In the graph:

The given locations are:

• Owen's House, A(11,3)

,

• David's House, B(15,13)

,

• School, C(3,18)

We determine both Owen's and David's distance from the school using the distance formula.

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Owen's distance from school (AC)

\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}

David's distance from school (BC)

\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}

We see from the calculations that David lives closer to the school, and by 4 miles.

The graph below is attached for further understanding:

5 0
1 year ago
Can someone please help its due soon.ill give brainliest!
leva [86]

Answer:

18 and 6

4x= 18+6

4x=24

x=24/4

x=6

4x= 6×4

24

2x+3

6×2

12+3

15

3 0
3 years ago
-9(m + 2) + 4(6 – 7m)
Travka [436]
I think it is -37m+17 but I am not sure
4 0
3 years ago
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