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il63 [147K]
3 years ago
14

BRAINLIEST TO FIRST RIGHT ANSWER The coefficients of the first three terms in the expansion of (x – y) 4 are a) 1, –4, –6 b) 1,

–4, 6 c) 1, 4, 6 d) 1, 3, 5
Mathematics
1 answer:
Zigmanuir [339]3 years ago
7 0

Answer:

b) 1, -4, 6

Step-by-step explanation:

(x-y)^4=

(x-y)(x-y)(x-y)(x-y)=

(x^2-xy-xy-y^2)(x^2-xy-xy-y^2)=

(x^2-2xy-y^2)(x^2-2xy-y^2)=

x^4-4x^3y+6x^2y^2-4xy^3+y^4

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Rewrite the equation so that it does not have fractions
olya-2409 [2.1K]

Answer:

\frac{2 - 3}{4x}  =   \frac{5}{12}  \\ 4x(5) = 12(2 - 3) \\ 20x = 24 - 36 \\ 20x = 12

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2 years ago
Need help plz fast I do anything
Kipish [7]

Answer:

which ones?

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
(42) A school only provides bus service
Nostrana [21]

Answer:

~1.8 mile

Step-by-step explanation:

Michael lives at the closest  point to the school (the origin) on Maple Street,  which can be represented by the line  y = 2x – 4.

This means Michael's house will be the intersection point of line y1 (y = 2x - 4) and line y2 that is perpendicular to y1 and passes the origin.

Denote equation of y2 is y = ax + b,

with a is equal to negative reciprocal of 2 => a = -1/2

y2 pass the origin (0, 0) => b = 0

=> Equation of y2:

y = (-1/2)x

To find location of Michael's house, we get y1 = y2 or:

     2x - 4 = (-1/2)x

<=> 4x - 8 = -x

<=> 5x = 8

<=> x = 8/5

=> y = (-1/2)x = (-1/2)(8/5) = -4/5

=> Location of Michael' house: (x, y) = (8/5, -4/5)

Distance from Michael's house to school is:

D = sqrt(x^2 + y^2) = sqrt[(8/5)^2 + (-4/5)^2) =  ~1.8 (mile)

4 0
3 years ago
On August 2nd, 1988, a US District Court judge imposed a fine on the city of Yonkers, New York, for defying a federal court orde
Minchanka [31]

Answer:

a)

A(t) = 1,000,000 + 10,000,000t

B(t)= 0.01 +0.02t

b) No.

Step-by-step explanation:

a)<u> Penalty A:</u> 1 million dollars on August 2 and the fine increases by 10 million dollars each day thereafter.

If t represents the number of days after August 2,

A(t) = 1,000,000 + 10,000,000t

<u>Penalty B</u>: 1 cent on August 2 and the fine doubles each day thereafter.

A(t) = 0.01 + 2t(0.01) = 0.01 + 0.02t

b) Assuming your formulas in part (a) hold for t≥0, is there a time such that the fines incurred under both penalties are equal?

To solve this, we would have to equal both formulas and solve for t.

1,000,000 + 10,000,000t=0.01+0.02t\\9,999,999.98t=-999,999.99

By taking a look at this equation, we see that when we solve for t, t will be a negative number. Since the formulas are valid for t≥0, we can conclude that there won't be a time such that the fines are equal.

4 0
3 years ago
Version A
trapecia [35]
The answer should be c
3 0
4 years ago
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