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serious [3.7K]
4 years ago
14

What's the unit cube

Mathematics
2 answers:
olya-2409 [2.1K]4 years ago
6 0

more formally a cube of side 1 cube is a cube whose sides are one unit long
Effectus [21]4 years ago
3 0
A unit cube is just a cube just like this one:

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Which value for x makes the sentence true? 12x-5 = 5x + 23<br><br> A. 6<br> B. 9<br> C. 4 <br> D. -1
hammer [34]

Answer:

C

Step-by-step explanation:

If you plug in 4 to the equation it looks like this:

48-5= 20+23. Subtract 5 and add 23, to get 43=43. That statement is true so it's 4.

3 0
3 years ago
Find the unknow based length of triangle A=31.5 in 2 H=7 in then wat is Abe
Sauron [17]

Answer:

the answer is 9

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
What are all the numbers with the absolute value of -4​
KIM [24]

Answer:

Just 4 and -4

4 0
3 years ago
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Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)
DENIUS [597]

Compute the necessary values/derivatives of f(x) at x=1:

f(1)=3

f'(1)=2

f''(1)=-12

f'''(1)=-12

f^{(4)}(1)=48

f^{(5)}(1)=120

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) f(x) at x=1 by

T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5

T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5

###

Another way of doing this would be to solve for the coefficients a,b,c,d,e,g in

f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5

by expanding the right hand side and matching up terms with the same power of x.

5 0
3 years ago
Help picture attached!!
tigry1 [53]
I think it would be D, because you would divide the 1/8 be 5, giving you 1/40
6 0
3 years ago
Read 2 more answers
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