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3 years ago

Dollar tree math questions

Mathematics

2 answers:

Korvikt [17]3 years ago

6 0

$$ Haha but where are the questions tho?

madam [21]3 years ago

3 0

Answer:

question?

Step-by-step explanation:

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What are the constants in this expression?<br> -10.6+9\10+2\5m-2.4n+3m

Annette [7]

-10.6, and 9/10, are the constants

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3 years ago

How would I write that translation rule (right 2, down 10)

Nady [450]

Answer: (2, -10)

Down and left are negative right and up are positive.

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2 years ago

Find an equation of the circle that has center (-1, 3) and passes through (1.<br> 1)

pshichka [43]

Answer:

(x + 1)² + (y - 3)² = 8

Step-by-step explanation:

radius (r): √(1 - (-1))² + (1 - 3)² = √8 = 2√2

Formula for circle with center (h,k): (x – h)² + (y – k)² = r²

(x + 1)² + (y - 3)² = 8

3 0

3 years ago

Read 2 more answers

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

9/8 divided by 3/8<br><br> 3/5 divided by 2/3

suter [353]

9/8 divided by 3/8 is 3 (because of the same denominator look at the first number)

3/5 divided by 2/3 is 0.9 hope this helps

8 0

2 years ago