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Monica [59]
3 years ago
15

Hey Guys ,Help Meh _/\_​

Mathematics
2 answers:
Ivan3 years ago
4 0

Answer:

6 km is the least than to request in math

Step-by-step explanation:

the math is than for the meaning of the numbers of math

USPshnik [31]3 years ago
4 0
<h2>Answer:</h2>

We know that,

<u>LSA of rectangular prism = (a + b + c + d)l</u>

= (4 + 7 + 11 + 7)6

= 29 × 6

= 174km².

<u>Correct choice</u> - [D] 174km².

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Ask Your Teacher The circumference of a sphere was measured to be 90 cm with a possible error of 0.5 cm. (a) Use differentials t
SCORPION-xisa [38]

Answer:  

a)  28,662 cm²  max error

    0,0111     relative error

b) 102,692 cm³  max error

   0,004     relative error

   

Step-by-step explanation:

Length of cicumference is: 90 cm

L = 2*π*r

Applying differentiation on both sides f the equation

dL  =  2*π* dr    ⇒  dr = 0,5 / 2*π

dr =  1/4π

The equation for the volume of the sphere is  

V(s) =  4/3*π*r³     and for the surface area is

S(s) = 4*π*r²

Differentiating

a) dS(s)  =  4*2*π*r* dr    ⇒  where  2*π*r = L = 90

Then    

dS(s)  =  4*90 (1/4*π)

dS(s) = 28.662 cm²   ( Maximum error since dr = (1/4π) is maximum error

For relative error

DS´(s)  =  (90/π) / 4*π*r²

DS´(s)  = 90 / 4*π*(L/2*π)²      ⇒   DS(s)  = 2 /180

DS´(s) = 0,0111 cm²

b) V(s) = 4/3*π*r³

Differentiating we get:

DV(s) =  4*π*r² dr

Maximum error

DV(s) =  4*π*r² ( 1/  4*π*)   ⇒  DV(s) = (90)² / 8*π²

DV(s)  =  102,692 cm³   max error

Relative error

DV´(v) =  (90)² / 8*π²/ 4/3*π*r³

DV´(v) = 1/240

DV´(v) =  0,004

3 0
3 years ago
Read 2 more answers
Please help me answer these. Its all about upper and lower bounds.
sergeinik [125]

Answer:

red question= 8.5 yellow question=4.25

Step-by-step explanation:

5 0
3 years ago
What are the steps to(2x+3y)4
lilavasa [31]
(2x + 3y)4

Distribute 

4*2x = 8x
4*3y = 12y

8x + 12y 

Answer: 8x + 12y
8 0
3 years ago
Read 2 more answers
Find the domain of the function y = 3 tan(23x)
solmaris [256]

Answer:

\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

In other words, the x in f(x) = 3\, \tan(23\, x) could be any real number as long as x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) for all integer k (including negative integers.)

Step-by-step explanation:

The tangent function y = \tan(x) has a real value for real inputs x as long as the input x \ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

Hence, the domain of the original tangent function is \mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

On the other hand, in the function f(x) = 3\, \tan(23\, x), the input to the tangent function is replaced with (23\, x).

The transformed tangent function \tan(23\, x) would have a real value as long as its input (23\, x) ensures that 23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

In other words, \tan(23\, x) would have a real value as long as x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right).

Accordingly, the domain of f(x) = 3\, \tan(23\, x) would be \mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

4 0
2 years ago
Right triangle with angles 12,8 and y Pythagorean theorem
igomit [66]

Answer:

Step-by-step explanation:

4 0
3 years ago
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