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3 years ago

Pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeaeeeeeeeeeee helpepppp meeeeeeeeeeeeeeeeeeeeeeee

Mathematics

1 answer:

Likurg_2 [28]3 years ago

4 0

Answer : Option D

Step by step explanation :
There is a 50-50 chance of snowing in Hawaii.

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The height of a soccer ball that is kicked from the ground can be approximated by the function:

kumpel [21]

Why do you want me a little bit of

3 0

3 years ago

Anna made 20/28 free throws in basketball this season.To the nearest thousands,what is her free-throw average?

Afina-wow [57]

Answer:

0.714 free throws.

Step-by-step explanation:

Anna made 20 out of 28 free throws.

Free-throw average = $\frac{20}{28}$ = 0.7142857143 free throws.

This equals 0.714 free throws (answer rounded up to the nearest thousandths)

5 0

4 years ago

Aya = a1 + (n - 1)d

Alenkasestr [34]

Answer:

The 26th term of an arithmetic sequence is:

$a_{26}=67$

Hence, option A is true.

Step-by-step explanation:

Given

a₁ = -33

d = 4

An arithmetic sequence has a constant difference 'd' and is defined by

$a_n=a_1+\left(n-1\right)d$

substituting a₁ = -33 and d = 4 in the nth term of the sequence

$a_n=-33+\left(n-1\right)4$

$\:a_n=-33+4n-4$

$a_n=4n-37$

Thus, the nth term of the sequence is:

$a_n=4n-37$

now substituting n = 26 in the nth term to determine the 26th term of the sequence

$a_n=4n-37$

$a_{26}=4\left(26\right)-37$

$a_{26}=104-37$

$a_{26}=67$

Therefore, the 26th term of an arithmetic sequence is:

$a_{26}=67$

Hence, option A is true.

7 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

katie received 15 gifts and a total of $100 in cash for her birthday. She buys one book that costs $23, and she wants to donate

Lady bird [3.3K]

Answer:

Katie will have 7 dollars left over.

Step-by-step explanation:

100

She buys a book so subtract 23 from 100.

100-23= 77

Katie has 77 dollars and wants to donate $10 to as much charities. 10 can go into 77, 7 times, which means she can only donate to 7 charities in $10.

6 0

3 years ago