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2 years ago

I i rlly don't know this gets harder tho

Mathematics

1 answer:

laiz [17]2 years ago

5 0

Answer:

Step-by-step explanation:

4(n + 10) = 34

4n + 40 = 34

If you need the rest...

4n = -6

n = -6/4

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In fraction form, what is 3/5÷1/3?<br><br>Thank you if answered

sineoko [7]

9/5 is in fraction form

6 0

3 years ago

Read 2 more answers

I need help on this question for the beginning of the year test

nika2105 [10]

Answer:

She still has to cut 1 3 /16

Step-by-step explanation:

Olivia wants to cut 3 3/4 inches from a piece of string. Converting 3 3/4 to improper fraction, it becomes 15/4 inches.

She already cut off 2 9/16 inches converting to improper fractions it becomes 41/16 inches.

15/4 - 41/16 = (60-41)/16 = 19/16 inches

19/16 = 1.187 ~ 1

19 - (16 x 1) =3

8 0

3 years ago

What Fraction Of The Wall Did Kevin Paint?

Mila [183]

Kevin painted $\frac{1}{9}$ of the wall.

Solution:

Fraction of the wall painted by Elena = $\frac{5}{9}$

Fraction of the wall painted by Matthew = $\frac{3}{9}$

Fraction of the wall painted by Kevin = ?

Full wall can be taken as 1.

<u>To find the wall painted by Kevin:</u>

Rest of the wall = Full wall – painted by Elena – Painted by Matthew

$$=1-\frac{5}{9}-\frac{3}{9}$

$$=\frac{1}{1} -\frac{5}{9}-\frac{3}{9}$

To make the denominator same, multiply the numerator and denominator of the first term by 9.

$$=\frac{9}{9} -\frac{5}{9}-\frac{3}{9}$

$$=\frac{9-5-3}{9}$

$$=\frac{1}{9}$

Rest of the wall = $\frac{1}{9}$

Hence the kevin painted $\frac{1}{9}$ of the wall.

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 $-4a^3b$ .

3 0

3 years ago

true or false: when multiplying radicals, you multiply the radicands together and the numbers outside the radicals together

Dmitrij [34]

You can only multiply numbers that are inside the radical symbols

7 0

3 years ago