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2 years ago

8

Devin is on his school's swim team. During his 2-week winter break, he swims 30 laps each week to stay in shape. If each lap is

25 yards, how many feet does he swim during his winter break?

1 answer:

tamaranim1 [39]2 years ago

7 0

I think... 1500

Step-by-step explanation:

multiply 30 and 25 then times 750 as your answer by 2 to get 1500

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17. If a + b = c, which of the following statements is true?

Scilla [17]

Answer:

c: c-a = b

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2 years ago

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A hat contains four balls. The balls are numbered 2, 4, 4, and 7. One ball is randomly selected and

vfiekz [6]

Answer:

If the sum is 6, a small cone will be ordered.
If the sum is 8 or 9, a medium cone will be ordered.
If the sum is 11, a large cone will be ordered.

Step-by-step explanation:

The table below shows the possible outcomes from drawing two balls. We see the relative frequencies to be ...

6: 4

8: 2

9: 2

11: 4

To make a fair decision, we want the relative frequencies of the decision possibilities to be the same. If we combine sums 8 and 9 to one category (medium cone, for example), then 6, (8 or 9), and 11 will all have the same relative frequency: 4.

The appropriate "fair decision" choice is ...

6: small cone

8 or 9: medium cone

11: large cone

6 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 $-4a^3b$ .

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2 years ago

5 and 1/4 divided by negative 6 simplify

Zanzabum

Answer:

-0.875

Step-by-step explanation:

5 1/4 = 5.25

5.25/ -6 = -0.875

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Pls help<br> Spam answers will be reported<br> Give a good explanation

Pavlova-9 [17]

Answer:

X = 20

The picture explains my answer

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