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Misha Larkins [42]
3 years ago
5

Hideki buys 2 sweaters for $18 each. He can use one of the following coupons:

Mathematics
2 answers:
Marizza181 [45]3 years ago
6 0
Coupon 2 equals 27 dollar coupon one equals = 28.80. You will get the greatest discount off of coupon 1 because you pay $1.80 less. What i did was i used the calculator
jeka943 years ago
3 0
Answer:

$1.80 less!!
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Please help me with my math! Thank you!
tangare [24]

Answer:

c determine the side length of one tablecloth correct to the nearest tenth of a meter

3 0
3 years ago
Someonee can u tell ne​
Vesnalui [34]

Answer:

35

Step-by-step explanation:

i did the sqrt of 1225 and got 35 so there should be 35 rows each with 25 plants

4 0
3 years ago
What is the value of (–7 + 3i) – (2 – 6i)
Sidana [21]
((–7) + (3 + i)) - (2 - (6 + i)) =-9 + 9 i
3 0
3 years ago
John is making himself a lunch. He has 3 different soups to choose from, 4 kinds of pop, and 3 different kinds of fruit. How man
notsponge [240]
36 combinations im pretty sure 3*4*3=36
5 0
3 years ago
Expand the following using the Binomial Theorem and Pascal’s triangle. (x + 2)6 (x − 4)4 (2x + 3)5 (2x − 3y)4 In the expansion o
ivolga24 [154]
\bf (2x+3)^5\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2x)^5(+3)^0\\
2&+5&(2x)^4(+3)^1\\
3&+10&(2x)^3(+3)^2\\
4&+10&(2x)^2(+3)^3\\
5&+5&(2x)^1(+3)^4\\
6&+1&(2x)^0(+3)^5
\end{array}

as you can see, the terms exponents, for the first term, starts at highest, 5 in this case, then every element it goes down by 1, till it gets to 0

for the second term, starts at 0, and every element it goes up by 1, till it gets to the highest

now, to get the coefficient, they way I get it, is "the product of the current coefficient and the exponent of the first term, divided by the exponent of the second term plus 1"

notice the first coefficient is always 1

so...how did we get 10 for the 3rd element?  well, 5*4/2
how did we get 10 for the fourth element?  well, 10*2/4


\bf (2x-3y)^4\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2x)^4(-3y)^0\\
2&+4&(2x)^3(-3y)^1\\
3&+6&(2x)^2(-3y)^2\\
4&+4&(2x)^1(-3y)^3\\
5&+1&(2x)^0(-3y)^4
\end{array}


\bf (3a+4b)^8\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(3a)^8(+4b)^0\\
2&+8&(3a)^7(+4b)^1\\
3&+28&(3a)^6(+4b)^2\\
4&+56&(3a)^5(+4b)^3\\
5&+70&(3a)^4(+4b)^4\\
6&+56&(3a)^3(+4b)^5\\
7&+28&(3a)^2(+4b)^6\\
8&+8&(3a)^1(+4b)^7\\
9&+1&(3a)^0(+4b)^8
\end{array}

and from there, you can simplify the elements of the expansion by combining the coefficients

like for example, the 7th element of (3a+4b)⁸ will then be 1032192a²b⁶
7 0
3 years ago
Read 2 more answers
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