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V125BC [204]
3 years ago
5

What's the square root of 30(.8)(90)??

Mathematics
1 answer:
Burka [1]3 years ago
3 0
<span>square root of 30(.8)(90)
</span>√30(.8)(90)
√2700(.8)
√2160
12√15

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When a figure is translated, reflected, or rotated, what's the same about the original igure and its image? What's different?
Liula [17]

Answer:

The image of a translation, reflection, or rotation is congruent to the original figure, and the image of a dilation is similar to the original figure. Two figures are similar when one can be obtained from the other by a sequence of translations, reflections, rotations, and dilations.

7 0
3 years ago
MATH HELP PLEASE!!! PLEASE HELP!!!
grandymaker [24]
The answer is:  [C]:  " f(c) = \frac{9}{5} c  + 32 " .
________________________________________________________

Explanation:

________________________________________________________
Given the original function:  

" c(y) = (5/9) (x <span>− 32) " ; in which "x = f" ; and "y = c(f) " ;
________________________________________________________
</span>→  <span>Write the original function as:  " y = </span>(5/9) (x − 32) " ; 

Now, change the "y" to an "x" ; and the "x" to a "y"; and rewrite; as follows:
________________________________________________________
    x = (5/9) (y − 32) ; 

Now, rewrite THIS equation; by solving for "y" ; in terms of "x" ; 
_____________________________________________________
→ That is, solve this equation for "y" ; with "c" as an "isolated variable" on the
 "left-hand side" of the equation:

We have:

→  x  =  " (  \frac{5}{9}  ) * (y − 32) " ;

Let us simplify the "right-hand side" of the equation:
_____________________________________________________

Note the "distributive property" of multiplication:
__________________________________________
a(b + c) = ab + ac ;  <u><em>AND</em></u>:

a(b – c) = ab – ac
.
__________________________________________

As such:
__________________________________________

" (\frac{5}{9}) * (y − 32) " ; 

=  [ (\frac{5}{9}) * y ]   −  [ (\frac{5}{9}) * (32) ] ; 


=  [ (\frac{5}{9}) y ]  − [ (\frac{5}{9}) * (\frac{32}{1})" ;

=  [ (\frac{5}{9}) y ]  − [ (\frac{(5*32)}{(9*1)} ] ; 

=  [ (\frac{5}{9}) y ]  −  [ (\frac{(160)}{(9)} ] ; 

= [ (\frac{5y}{9}) ]  −  [ (\frac{(160)}{(9)} ] ; 

= [ \frac{(5y-160)}{9} ] ;  
_______________________________________________
And rewrite as:  

→  " x  =  \frac{(5y-160)}{9} "  ;

We want to rewrite this; solving for "y";  with "y" isolated as a "single variable" on the "left-hand side" of the equation ;

We have:

→  " x  =  \frac{(5y-160)}{9} "  ; 

↔  " \frac{(5y-160)}{9} = x ; 

Multiply both sides of the equation by "9" ; 

 9 * \frac{(5y-160)}{9}  =  x * 9 ; 

to get:

→  5y − 160 = 9x ; 

Now, add "160" to each side of the equation; as follows:
_______________________________________________________

→  5y − 160 + 160 = 9x + 160 ; 

to get:

→  5y  =  9x + 160 ; 

Now,  divided Each side of the equation by "5" ; 
      to isolate "y" on one side of the equation; & to solve for "y" ; 

→  5y / 5  = (9y + 160) / 5 ; 

to get: 
 
→  y = (9/5)x + (160/5) ; 

→  y =  (9/5)x + 32 ; 

 →  Now, remember we had substituted:  "y" for "c(f)" ; 

Now that we have the "equation for the inverse" ;
     →  which is:  " (9/5)x  + 32" ; 

Remember that for the original ("non-inverse" equation);  "y" was used in place of "c(f)" .  We have the "inverse equation";  so we can denote this "inverse function" ; that is, the "inverse" of "c(f)" as:  "f(c)" .

Note that "x = c" ; 
_____________________________________________________
So, the inverse function is: "  f(c) = (9/5) c  + 32 " .
_____________________________________________________

 The answer is:  " f(c) = \frac{9}{5} c  + 32 " ;
_____________________________________________________
 →  which is:  

→  Answer choice:  [C]:  " f(c) = \frac{9}{5} c  + 32 " .
_____________________________________________________
6 0
3 years ago
Solve for x 5(x+4)=2x+20+2x
spayn [35]
Hey!
so first distribute the 5 to x + 4
5x+20 = 2x + 20 + 2x
combine 2x and 2x
5x+20 = 4x + 20
Now you want the constant and variable on different sides so subtract 4x to the other side
5x - 4x + 20 = 20
1x + 20 = 20
now subtract the 20 on the 1x + 20 side to the other side
1x = 0
divide by 1
0 divided by anything is 0
So x = 0
Hope this helps!
8 0
3 years ago
Read 2 more answers
6X plus 3Y equals 3 3x minus Y equals four
Delvig [45]

Answer:

Step-by-step explanation:

6x + 3y = 3 simplify by dividing both sides by 3 2x + y = 1

3x - y = 4 the 1st equation has a positive y, the 2nd has a negative y. Use elimination

to reduce them down to a single variable equation:

2x + y = 1

3x - y = 4

________

5x = 5 solve

X = 1 substitute back in to one of the original equations to find y

3x - y = 4

3(1) - y = 4

- y = 1

y = -1 (1,-1)

8 0
3 years ago
Help........Write an equation and solve:
Tom [10]

Answer:

5 months

Step-by-step explanation:

Equate the formulas for the weights of the two boys:

J's weight = 120 lb + (10 lb/mo)m  = D's weight = 150 lb + (4 lb/mo)m

Solve as follows:  Subtract 120 lb from both sides:

(10 lb/mo)m = 30 lb + (4 lb/mo)m.

Then:  (6 lb/mo)m = 30 lb, and m = (30 lb) / (6 lb/mo) = 5 months

3 0
3 years ago
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