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Ilia_Sergeevich [38]
3 years ago
7

One of two complementary angles is 4 degrees more than the other. Find the angles. (Recall that complementary angles are angles

whose sum is 90 degrees.)
Which of the following equations can not be used to solve the problem if x represents one of the angles?
Mathematics
1 answer:
emmasim [6.3K]3 years ago
8 0
   
\displaystyle  \\ 
\alpha + \beta = 90^o \\ 
\alpha - \beta = 4^o \\ 
\texttt{----------- } +  \\ 
2\alpha ~~/~= 94^o \\  \\ 
\alpha  =  \frac{94}{2} = \boxed{47^o}\\  \\ 
 \beta = 90 -\alpha  = 90 - 47 = \boxed{43^o}




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Answer:

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Step-by-step explanation:

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During dance practice, Sasha drank 2 1/2 pints of water, and on the way home she drank 1/2 cup of water. How much did she drank
Anarel [89]

Answer: 5 cups 4 ounces

Step-by-step path to answer:

Pints to cups:

1 pint= 2 cups

2 pints= 4 cups

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3 0
3 years ago
Write an equation of a parabola that opens to the left, has a vertex at the origin, and a focus at (–9, 0.
Georgia [21]
The standard equation of parabola:

(y-k)²=4p(x-h), with:

a) vertex = (h,k)

b) focus = (h+p, k)

c) directrix = (x=h-p)


Since this parabola has a vertex at (0,0) that means h=k=0

Hence the equation becomes: y²=4px, let's calculate p:
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What type of graph would have the title, "Mile Time of a Typical 5th Grade Student
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aivan3 [116]

Answer:

We verified that a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]

Hence proved

Step-by-step explanation:

Given equation is a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]

We have to prove that a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]

That is to prove that LHS=RHS

Now taking RHS

\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]

=\frac{a+b+c}{2}[a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2]  (using (a-b)^2=a^2-2ab+b^2)

=\frac{a+b+c}{2}[2a^2-2ab+2b^2-2bc+2c^2-2ac]  (adding the like terms)

=\frac{a+b+c}{2}[2a^2+2b^2+2c^2-2ab-2bc-2ac]

=\frac{a+b+c}{2}\times 2[a^2+b^2+c^2-ab-bc-ac]

=a+b+c[a^2+b^2+c^2-ab-bc-ac]

Now multiply the each term to another each term in the factor

=a^3+ab^2+ac^2-a^2b-abc-a^2c+ba62+b^3+bc^2-ab^2-b^2c-abc+ca^2+cb^2+c^3-abc-bc^2-ac^2]

=a^3+b^3+c^3-3abc (adding the like terms and other terms getting cancelled)

=a^3+b^3+c^3-3abc =LHS

Therefore LHS=RHS

Therefore a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]

Hence proved.

8 0
4 years ago
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