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jek_recluse [69]
2 years ago
13

LOOK BELOW AT THE PICTURE

Mathematics
2 answers:
Tems11 [23]2 years ago
8 0

Answer:

C. 1796.08 in^2

Step-by-step explanation:

Area of circle is \pi r^2, r is the radius

remember radius is from the center to the edge, so it is half the diameter

now we need to find the area of the large circle (table) minus the small circle (hole)

the radius of the large circle is 48/2 which is 24

the area is \pi 24^2 replacing r in the formula with 24, the radius. Using 3.14 as pi, the answer is 1808.64

the radius of the small circle is 4/2 which is 2

the area is \pi 2^2 so the calculated value would be 12.56

now subtract the area of the hole (small circle) from the area of the table (large circle)

1808.64 - 12.56 = 1796.08

and there's your answer

hope that helps, lmk if it doesn't :)

dolphi86 [110]2 years ago
6 0

Answer:

Step-by-step explanation:

r₁ = radius of tabletop = 24 in

r₂ = radius of hole = 2 in

area of hole = πr₂² = 4π in²

area of tabletop = πr₁² - πr₂² = 576π - 4π = 572π ≈ 1796.08 in²

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Simplify the function. Than determine the key aspects of the function
Tom [10]

To simplify the function, we need to know some basic identities involving exponents.


1. b^(ax)=(b^x)^a=(b^a)^x

2. b^(x/d) = (b^x)^(1/d) = ((b^(1/d)^x)


Now simplify f(x), where

f(x)=(1/3)*(81)^(3*x/4)

=(1/3)(3^4)^(3*x/4) [ 81=3^4 ]

=(1/3)(3^(4*3*x/4) [ rule 1 above ]

=(1/3) (3^(3*x)

=(1/3)(3^(3x)) [ or (1/3)(27^x), by rule 1 ]



(A) Initial value is the value of the function when x=0, i.e.

initial value

= f(0)

=(1/3)(3^(3x))

=(1/3)(3^(3*0))

=(1/3)(3^0)

=(1/3)(1)

=1/3


(B) the simplified base base is 3 (or 27 if the other form is used)


(C) The domain for an exponential function is all real values ( - ∞ , + ∞ ).


(D) The range of an exponential function with a positive coefficient and without vertical shift is ( 0, + ∞ ).

8 0
3 years ago
Which of the following statements are true? Select all that apply.
Gnesinka [82]

Answer:

a and c

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Let D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36} Determine which of the following statements are true and which are false. a)
morpeh [17]

Answer:

\forall x\in D if x is odd then x> 0 is true statement.

\forall x\in D if x is odd then x> 0 is true statement.

\forall x\in D if x is even then x≤0 is false statement.

\forall x\in D If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.

\forall x\in D if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.

Step-by-step explanation:

Consider the provided information.

D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36}

Part (A) \forall x\in D if x is odd then x> 0

Here only even numbers are less than 0 that means the statement is true.

\forall x\in D if x is odd then x> 0 is true statement.

Part (B) \forall x\in D if x is less than 0 then x is even.

Here only even numbers are less than 0 that means the statement is true.

\forall x\in D if x is odd then x> 0 is true statement.

Part (C) \forall x\in D if x is even then x≤0

Here we can see that 16, 26, 32, 36 are even number and also greater than 0. Thus the statement is false.

\forall x\in D if x is even then x≤0 is false statement.

Part (D) \forall x\in D If the ones digit of x is 2, then the tens digit is 3 or 4.

There is only one number whose ones digit is 2. i.e. 32 also the tens digit of the number 32 is 3. Which makes the above statement true.

\forall x\in D If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.

Part (E) \forall x\in D if the ones digit of x is 6, then the tens digit is 1 or 2.

Numbers having ones digit 6 are: 16, 26 and 36

Here, the tens digits are 1, 2 and 3 which is contradict to our statement. Hence the provided statement is false.

\forall x\in D if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.

8 0
3 years ago
Hey could anyone help me with this question :)
Marta_Voda [28]

Answer:

cuz i am batman hhhhhhhhhhhh

7 0
2 years ago
I need help on this one please
saw5 [17]

Answer:

B

Step-by-step explanation:

When looking at the number line, we can see the distance between A and B is one unit

A |-2| + |-1| = 2+1 =3    That is not 1 unit

B 2-1  =1  that is  1 unit

C  -2 + -1 = -3  that is not 1 unit

D  |-1| - (-2) = 1 + 2 =3  that is not 1 unit

Even though it says all expressions,  there is only 1 that shows a distance of 1 unit

6 0
3 years ago
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