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pav-90 [236]
3 years ago
9

Can someone help me please

Mathematics
2 answers:
maks197457 [2]3 years ago
8 0
<span>8y - 3y = 5y because you just subtract and then add y at the end</span>
ELEN [110]3 years ago
3 0
9. 5y
10.10x^2-1 
not done 
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(X+h)^3 + (x+h)-(x^3+x)
ICE Princess25 [194]
\dfrac{(x+h)^3 + (x+h)-(x^3+x)}{h}=\\&#10;\dfrac{x^3+3x^2h+3xh^2+h^3+x+h-x^3-x}{h}=\\&#10;\dfrac{3x^2h+3xh^2+h^3+h}{h}=\\3x^2+3xh+h^2+1
8 0
3 years ago
Please help me with this
ElenaW [278]

Answer:

Equation: (3+x) +x= 49

Lynette's age: 23

Josh's age: 26

7 0
3 years ago
Find the area of the shaded region. round answers to the nearest tenth. assume all inscribed polygons are regular. 29.3 units2
FromTheMoon [43]
The picture in the attached figure

we know that
the area of the shaded region is equal to
(2/3)*[area of the circle - the area of the triangle]  

step 1
Find the area of a circle Ac
Ac = π r²
Ac = π (6)²
Ac = 113.10 units²

step 2
find the area of the triangle At
The triangle is an equilateral triangle with angles on each corner equal to 60 degrees. Meanwhile,
the 3 angles at the center is 120 degrees each since a circle is 360 degree.
 We know that the radius (line from centerpoint to corner) is equivalent to 6.
Using the cosine law,
we can calculate for the length of one side.
s² = 6^ + 6² – 2 (6) (6) cos 120
s² = 108
s = 10.4 units
Since this is an equilateral triangle, therefore, all sides are equal.
The area for this is:
At = (sqrt3 / 4) * s²
At = 46.77 units²

step 3
the area of the shaded region=(2/3)*[area of the circle - the area of the triangle]  
the area of the shaded region=(2/3)*[113.10-46.77]------> 44.22 units²

therefore

the answer is
the area of the shaded region is 44.22 units²


5 0
3 years ago
Doug says that this clock shows 8:43.is he correct? Explain why or why not.
ss7ja [257]
Is this a joke because it made me laugh
5 0
3 years ago
Read 2 more answers
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
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