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Pavlova-9 [17]
2 years ago
7

Pls help me first get 10 you know

Mathematics
1 answer:
Luba_88 [7]2 years ago
7 0

Option D, 63 cm

Explanation:

The scale of model is of local park  is

This means that each dimension of the model is 21 times smaller than the actual park (to be developed in future)

Given -

Depth of the pond  centi meter

Depth of the real pond will be 21 times of the depth of the model.

Thus , real pond's depth is equal to

cm

Hence, option D is correct

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Laws Of Indicies<br>Can someone please help me with this?​
nasty-shy [4]

Answer:

See below

Step-by-step explanation:

1) :  {a}^{5}  {b}^{4}  {c}^{3}  \div  {a}^{2}  {b}^{3} c \\  \\  = {a}^{5 - 2}  {b}^{4 - 3}  {c}^{3 - 1}  \\  \\  = {a}^{3}  {b}^{1}  {c}^{2}  \\  \\ = {a}^{3}  {b}  {c}^{2}  \\  \\  \\2)\:  (x^2 y^3)^3\\\\= x^{2\times 3}y^{3\times 3}\\\|= x^{6}y^{9}\\\\\\3)\:5 {a}^{4}   \div ( {a}^{2}  \times 3a) \\  \\  = 5 {a}^{4}   \div (3 {a}^{2 + 1}  ) \\  \\  = 5 {a}^{4}   \div (3 {a}^{3}  ) \\  \\  =   \frac{5}{3}  {a}^{4 - 3}  \\  \\ =   \frac{5}{3}  {a} \\  \\  \\ 4)\: \frac{14 {a}^{5} }{2 {a}^{3} \times 7 {a}^{4}  }  \\  \\  =   \frac{14 {a}^{5} }{2 \times 7 {a}^{3 + 4} }   \\  \\  =   \frac{ \cancel{14} {a}^{5} }{ \cancel{14}{a}^{3 + 4} }    \\  \\  =   \frac{  {a}^{5} }{ {a}^{7} }   \\  \\  =  {a}^{5 - 7}  \\  \\  =  {a}^{ - 2}  \\  \\  =  \frac{1}{{a}^{2} }  \\  \\

8 0
3 years ago
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A culture started with 4,000 bacteria. After 8 hours, it grew to 4,800 bacteria. Predict how many bacteria will be present after
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Pavlova-9 [17]
What is the length of a soccer field.
6 0
3 years ago
Simplify u^2+3u/u^2-9<br> A.u/u-3, =/ -3, and u=/3<br> B. u/u-3, u=/-3
VashaNatasha [74]
  The correct answer is:  Answer choice:  [A]:
__________________________________________________________
→  "\frac{u}{u-3} " ;  " { u \neq ± 3 } " ; 

          →  or, write as:  " u / (u − 3) " ;  {" u ≠ 3 "}  AND:  {" u ≠ -3 "} ; 
__________________________________________________________
Explanation:
__________________________________________________________
 We are asked to simplify:
  
  \frac{(u^2+3u)}{(u^2-9)} ;  


Note that the "numerator" —which is:  "(u² + 3u)" — can be factored into:
                                                      →  " u(u + 3) " ;

And that the "denominator" —which is:  "(u² − 9)" — can be factored into:
                                                      →   "(u − 3) (u + 3)" ;
___________________________________________________________
Let us rewrite as:
___________________________________________________________

→    \frac{u(u+3)}{(u-3)(u+3)}  ;

___________________________________________________________

→  We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ;  since:

" \frac{(u+3)}{(u+3)} = 1 "  ;

→  And we have:
_________________________________________________________

→  " \frac{u}{u-3} " ;   that is:  " u / (u − 3) " ;  { u\neq 3 } .
                                                                                and:  { u\neq-3 } .

→ which is:  "Answer choice:  [A] " .
_________________________________________________________

NOTE:  The "denominator" cannot equal "0" ; since one cannot "divide by "0" ; 

and if the denominator is "(u − 3)" ;  the denominator equals "0" when "u = -3" ;  as such:

"u\neq3" ; 

→ Note:  To solve:  "u + 3 = 0" ; 

 Subtract "3" from each side of the equation; 

                       →  " u + 3 − 3 = 0 − 3 " ; 

                       → u =  -3 (when the "denominator" equals "0") ; 
 
                       → As such:  " u \neq -3 " ; 

Furthermore, consider the initial (unsimplified) given expression:

→  \frac{(u^2+3u)}{(u^2-9)} ;  

Note:  The denominator is:  "(u²  − 9)" . 

The "denominator" cannot be "0" ; because one cannot "divide" by "0" ; 

As such, solve for values of "u" when the "denominator" equals "0" ; that is:
_______________________________________________________ 

→  " u² − 9 = 0 " ; 

 →  Add "9" to each side of the equation ; 

 →  u² − 9 + 9 = 0 + 9 ; 

 →  u² = 9 ; 

Take the square root of each side of the equation; 
 to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ; 

→ √(u²) = √9 ; 

→ | u | = 3 ; 

→  " u = 3" ; AND;  "u = -3 " ; 

We already have:  "u = -3" (a value at which the "denominator equals "0") ; 

We now have "u = 3" ; as a value at which the "denominator equals "0"); 

→ As such: " u\neq 3" ; "u \neq -3 " ;  

or, write as:  " { u \neq ± 3 } " .

_________________________________________________________
6 0
3 years ago
A) 0.158<br> B) 15.8<br> C) 1.58<br> D) 1.625
vesna_86 [32]
D

1/8 = 0.125
0.125*5=0.625
0.625+1=1 5/8
6 0
3 years ago
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