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3 years ago

6

Determine for each number whether it is a rational number or irrational number

2 answers:

erastova [34]3 years ago

5 0

Picture is too blurry man.

shutvik [7]3 years ago

3 0

I cannot see this. try taking the picture again

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WORTH 100 POINTS!!!!!

aleksklad [387]

Answer:

1. This graph is not a function, because for every y there are more than one x.

2. This graph is not linear because it is not a straight line.

3. It took them 6 hours to get to their destination.

4. The destination was 325 miles away.

5. They might have stopped at a restaurant to eat.

6 0

2 years ago

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HELP DUE IN 10 MINS!

Scilla [17]

Answer:

$\checkmark \text{B. The area of the circle is } 729\pi,\\\checkmark \text{D. The arc length of the sector is } 12\pi$

Step-by-step explanation:

Area of a circle with radius $r$ : $r^2\pi$

Circumference of a sector with radius $r$ : $2r\pi$

Area of sector with angle $\theta$ : $r^2\pi\cdot \frac{\theta}{360}$ .

Arc length of sector with angle $\theta$ : $2r\pi\cdot \frac{\theta}{360}$

Using these equations, we get the following information:

Area of circle: $27^2\pi=729\pi$

Circumference of a circle: $2\cdot 27\cdot \pi=54\pi$

Area of sector: $27^2\pi\cdot \frac{80}{360}=162\pi$

Arc length of sector: $2\cdot 27\cdot \pi \cdot \frac{80}{360}=12\pi$

3 0

3 years ago

Harley took out a 1-year loan for $2800 at an electronics store to be paid back

WARRIOR [948]

Answer:

2919.25

Step-by-step explanation:

I took a risk and answer on my test, and failed but here y'all go

5 0

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 $-4a^3b$ .

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3 years ago

In a right triangle, one leg measures 4 inches and the other leg measures 6 inches. What is the length of the hypotenuse in inch

mina [271]

The length of the hypotenuse in inches is 2 square root 13

6 0

3 years ago

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