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Butoxors [25]
3 years ago
15

Please can anyone help me?

Mathematics
2 answers:
Nitella [24]3 years ago
8 0

Answer:

2 is the correct answer

hoa [83]3 years ago
6 0
i’m pretty sure the answer would be 1) linear :D
You might be interested in
Short Proofs - No Triangle Congruence
Marat540 [252]

Step-by-step explanation:

Given,

∠CBE ≅ ∠DBE ----> 1

∠CBE ≅ ∠BCA ----> 2

From 1 & 2,

∠BCA≅ ∠DBE ----> 3

∠DBE ≅ ∠BAC ----> 4

Reason,

Since BE II AC, corresponding angles are equal.

From 3 & 4,

∠BCA ≅ ∠BAC

BC is opposite to ∠BCA

AB is opposite to ∠ABC

Side opposite to congruent angles are always congruent.

Therefore,

AB ≅ BC

Hence proved.

3 0
2 years ago
I need help pretty plzzzzzz!!!!!!!!!!!!!
saveliy_v [14]

Answer:

Part a: yellow square 3 by 3 orange square 2 by 2

Part b: 78 inches is the area

Step-by-step explanation:

Hard to explain but its just the answer i think but part b is hard because i don't know how long the width is

8 0
3 years ago
A light source is located over the center of a circular table of diameter 4 feet. (See picture below) Find the height h of the l
Alex
Very nice to have an accompanied image!Illumination is proportional to the intensity of the source, inversely proportional to the distance squared, and to the sine of angle alpha.so that we can writeI(h)=K*sin(alpha)/s^2 ................(0)where K is a constant proportional to the light source, and a function of other factors.
Also, radius of the table is 4'/2=2', therefore, using Pythagoras theorem,s^2=h^2+2^2 ...........(1), and consequently,sin(alpha)=h/s=h/sqrt(h^2+2^2)..............(2)
Substitute (1) and (2) in (0), we can writeI(h)=K*(h/sqrt(h^2+4))/(h^2+4)=Kh/(h^2+4)^(3/2)
To get a maximum value of I, we equate the derivative of I (wrt alpha) to 0, orI'(h)=0or, after a few algebraic manipulations, I'(h)=K/(h^2+4)^(3/2)-(3*h^2*K)/(h^2+4)^(5/2)=K*sqrt(h^2+4)(2h^2-4)/(h^2+4)^3We see that I'(h)=0 if 2h^2-4=0, giving h=sqrt(4/2)=sqrt(2) feet above the table.
We know that I(h) is a minimum if h=0 (flat on the table) or h=infinity (very, very far away), so instinctively h=sqrt(2) must be a maximum.Mathematically, we can derive I'(h) to get I"(h) and check that I"(sqrt(2)) is negative (for a maximum).  If you wish, you could go ahead and find that I"(h)=(sqrt(h^2+4)*(6*h^3-36*h))/(h^2+4)^4, and find that the numerator equals -83.1K which is negative (denominator is always positive).
An alternative to showing that it is a maximum is to check the value of I(h) in the vicinity of h=sqrt(2), say I(sqrt(2) +/- 0.01)we findI(sqrt(2)-0.01)=0.0962218KI(sqrt(2))     =0.0962250K   (maximum)I(sqrt(2)+0.01)=0.0962218KIt is not mathematically rigorous, but it is reassuring, without all the tedious work.
3 0
3 years ago
Formar dos números pares de tres dígitos con el 2,4,8
Nitella [24]
248 , 842....aquí están los números....
8 0
3 years ago
Select all the expression that have the same value
Fantom [35]

Answer:

#2, 4, and 5 all have the same value of 64

Step-by-step explanation:

1) 16

3) 256

6) 16,777,216

5 0
3 years ago
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