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mash [69]
3 years ago
13

Write and solve an equation to find Marys. Mary is 6 years older than Ellen. Jim is 5 years older than Mary. If their total comb

ined age is 41. How old is Ellen?
Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

nggkkkb ghjkkjgh

ggjbdfhhkmbssrjokncfd

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Which is the graph for the equation y=-3x + 4?
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its look like this

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Which of the following is the equation of a line parallel to 3y=6x+5 that passes through (2,3)?
Verdich [7]

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The answer is C y= 2x-1

Step-by-step explanation:

7 0
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X+3y=3<br> 2x-4y=6<br><br> solve the equations
Gelneren [198K]
Subtract the second equation from double the first equation:

2x+6y-2x-(-4y)=0

10y=0

Therefore, y=0.  Substituting this into the first equation, we see that:

x+3(0)=3

x=3

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4 0
3 years ago
Read 2 more answers
) f) 1 + cot²a = cosec²a​
notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}

Another identity is:

\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}

Therefore:

\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}

Another identity:

\displaystyle \large{\sin^2x+\cos^2x=1}

Therefore:

\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}

Hence proved, this is proof by using identity helping to find the specific identity.

6 0
3 years ago
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