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ale4655 [162]
3 years ago
12

An irrational number is always almost or never a real number ????

Mathematics
2 answers:
SSSSS [86.1K]3 years ago
5 0

Answer:

The answer is: always

Step-by-step explanation:

All numbers even irrational numbers are real numbers.

Hope it helps!

OlgaM077 [116]3 years ago
3 0

Answer:

Always

Step-by-step explanation:

Irrational numbers are ALWAYS real numbers, no matter if they <em>repeat</em> or <em>terminate</em><em>.</em>

I am joyous to assist you anytime.

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Find a polynomial for the sum of the shaded<br> areas of the figure. A = 6, B = 4
soldi70 [24.7K]

Answer:

The polynomial for the sum of the shaded  \pi r² - 20 \pi

Step-by-step explanation:

Given as :

The figure is shown which is of concentric circle with radius B , A , r

The radius B = 4 unit

The radius A = 6 unit

Let The sum of shaded portion = x unit

Now, The circumference of circle = 2 \pi R , where R is the radius

So, for circle with radius B.

The circumference = 2 \pi R = 2 \pi B

Or, The circumference = 2 \pi × 4 = 8 \pi

<u>Similarly</u>

For circle with radius A.

The circumference = 2 \piR = 2 \pi A

Or, The circumference = 2 \pi × 6 = 12 \pi

Now, <u>The area of circle with radius r is</u>

Area = \pi ×radius × radius

Or, Area = \pi r²

Now,

The sum of shaded region area = The area of circle with radius r - ( The circumference with radius B + The circumference with radius A )

Or, The sum of shaded region area =  \pi r² - ( 8 \pi + 12 \pi )

Or,  The sum of shaded region area =  \pi r² - 20 \pi

Hence The polynomial for the sum of the shaded area is \pi r² - 20 \pi  Answer

5 0
4 years ago
Which answer correctly estimates the product of 79.623 × 4.83?
mamaluj [8]
The answer to this is 384.57909
6 0
3 years ago
PLEASE HELP ASAP! I don’t recall how to do this!
MakcuM [25]

Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

(1.05)^x=1.885

In order to solve for x, we have to get it out from its position of an exponent.  Do that by taking the natural log of both sides:

ln(1.05)^x=ln(1.885)

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

x=\frac{ln(1.885)}{ln(1.05)}

Do this on your calculator to find that

x = 12.99294297

For b. we will first apply the rule for "undoing" the addition of logs by multipllying:

ln(x*x^2)=5

Simplifying gives you

ln(x^3)=5

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

ln(x)=\frac{5}{3}

Take the inverse ln by raising each side to e:

e^{ln(x)}=e^{\frac{5}{3}}

The "e" and the ln on the left undo each other, leaving you with just x; and raising e to the power or 5/3 gives you that

x = 5.29449005

For c. begin by dividing both sides by 20 to get:

\frac{1}{2}=e^{.1x}

"Undo" that e by taking the ln of both sides:

ln(.5)=ln(e^{.1x})

When the ln and the e undo each other on the right you're left with just .1x; on the left we have, from our calculators:

-.6931471806 = .1x

x = -6.931471806

Question d. is a bit more complicated than the others.  Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:

(2^2)^x-6(2)^x=-8

Now we will bring over the -8 by adding:

(2^2)^x-6(2)^x+8=0

We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution.  Let's let u=2^x

When we do that, we can rewrite the polynomial as

u^2-6u+8=0

This factors very nicely into u = 4 and u = 2

But don't forget the substitution that we made earlier to make this easy to factor.  Now we have to put it back in:

2^x=4,2^x=2

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

2^2=2^x

Now that the bases are the same, we can say that

x = 2

For the second solution, we will raise the 2 on the right to a power of 1 to get:

2^x=2^1

Now that the bases are the same, we can say that

x = 1

5 0
3 years ago
.
GalinKa [24]
A. 515
b. 43
c. 3510
d. 1.9705
4 0
3 years ago
28 girls and 32 boys are playing a game. The coach will divide the girls and boys into groups that have girls and boys in each g
Sonja [21]

I believe that the coach can make 28 groups

7 0
4 years ago
Read 2 more answers
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