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sergey [27]
3 years ago
5

Y=33x+102 what is Y?

Mathematics
1 answer:
rusak2 [61]3 years ago
8 0

Answer:

you have y so i will guess u mean x your answer for x will be

Step-by-step explanation:

since 33 is ur first number and 11 evenly goes into 33 it will be

         Y                                         34

Y=    -----   SUBTRACTED BY     -------          

        33                                          11

<h2>Y EQUALS Y OVER 33 MINOUS 34OVER 11</h2>
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You have a 32-foot fence around a square garden. You paint 1/3 of one side of the fence. What fraction of the fence did you pain
dmitriy555 [2]
The answer is 1/12.

The perimeter of a square garden with side s is:
P = 4s

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Since it is painted 1/3 of 1/4 of the fence, the<span> fraction of the painted fence is:
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8 0
3 years ago
Please help me I really need help
Murljashka [212]

Answer:

x  = 77 degrees

Step-by-step explanation:

angle x  =  (76 + 78)/2 =   154/2 = 77 degrees

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3 0
3 years ago
Find the 2th term of the expansion of (a-b)^4.​
vladimir1956 [14]

The second term of the expansion is -4a^3b.

Solution:

Given expression:

(a-b)^4

To find the second term of the expansion.

(a-b)^4

Using Binomial theorem,

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here, a = a and b = –b

$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}

Substitute i = 0, we get

$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4

Substitute i = 1, we get

$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b

Substitute i = 2, we get

$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}

Substitute i = 3, we get

$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}

Substitute i = 4, we get

$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}

Therefore,

$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}

=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}

Hence the second term of the expansion is -4a^3b.

3 0
3 years ago
What is the factored form of x^2y^3 -2y^3 - 2x^2 +4
Law Incorporation [45]

(-x^2+4)

x^2y^3-2y^3-2x^2+4=x^2-2x^2+4=

(-x^2+4)

8 0
3 years ago
Rob is making a square pen for his dog to play in. Each
iogann1982 [59]

Answer:

A.900 square feet

Step-by-step explanation:

s×s=A

30×30=900

6 0
2 years ago
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