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2 years ago

What is the square root of 5 plus the square root of 5

Mathematics

1 answer:

Charra [1.4K]2 years ago

6 0

Answer:

That would be approximately 4.47.

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Reuben says, "two days ago i was 20 years old. later next year, i will be 23 years old." how is this possible?

Aneli [31]

Let us always base on the present age. We denote this by x. Now, the age he had 2 years ago would then be denoted as (x-2). Let's equate this to 20 years old.

x - 2 = 20
x = 20 + 2
x = 22 years old
He should be 22 years old now.

Let's check the other condition. After 1 year, his age should be (x+1). Let's equate this to 23 years old.

x + 1 = 23
x = 23 - 1
x = 22

Thus, this is possible if Reuben is 22 years old as of the present.

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3 years ago

I am about to get my homework done I just need these two. (Thank you if you answer!!)

AnnZ [28]

Answer:

$-7=h\\-6.81=q$

Step-by-step explanation:

For question #1:

$-3=h+8/2\\-3=h+4\\-7=h$

For question #2:

$q+16.41=9.6\\q=-6.81$

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1 year ago

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Juliette [100K]

How much after subtraction

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2 years ago

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Which two ratios form a proportion?

Alex17521 [72]

Step-by-step explanation:

4 and 5 is one 20/25

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3 years ago

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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

2 years ago

What is the square root of 5 plus the square root of 5​

What is the square root of 5 plus the square root of 5