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never [62]
3 years ago
6

Y=x+4 y=0 solve each system of equations by graphing

Mathematics
1 answer:
Svetach [21]3 years ago
5 0
Y=X+4,X∈R Thats what i got 
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Step-by-step explanation:

bx-3=y

bx-3+3=y+3

bx=y+3

bx/x=y+3/x

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A nuber to be multiplied is called
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Step-by-step explanation:

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3 years ago
1. Rewrite this relation as a function with the dependent variable on the left side of the equation. S+t2=16
Sauron [17]

Answer: d. s=16-t²

Step-by-step explanation:  the answer is s=16-t², because for this case is really important to keep in mind which is the value with the quadratic expression,  when we have a quadratic expression the graphic representation will be a parabola, in a parabola you will find a point that never change which is constant in the diagram.

also keep in mind that the values which are placed in the axis of the ordinates are the independent variables and in parabola the quadratic values are placed in the axis of the ordinates , and for this case is t² because of the quadratic expression, then we can find that t² is the independent variable and s is the dependent variable.

3 0
3 years ago
in 2009 a total of R36 000 was invested in two accounts. One account earned 7% annual interest and the other earned 9% .The tota
Akimi4 [234]

a = amount invested at 7%

b = amount invested at 9%

we know the amount invested was ₹36000, thus we know that whatever "a" and "b" are, a + b = 36000.  We can also say that

\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{7\% of a}}{\left( \cfrac{7}{100} \right)a}\implies 0.07a~\hfill \stackrel{\textit{9\% of b}}{\left( \cfrac{7}{100} \right)b}\implies 0.09b

since we know the interest earned from the invested was ₹2920, then we say that 0.07a + 0.09b = 2920.

\begin{cases} a + b = 36000\\\\ 0.07a+0.09b=2920 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{a + b = 36000\implies \underline{b = 36000-a}}~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.07a~~ + ~~0.09(\underline{36000-a})~~ = ~~2920} \\\\\\ 0.07a+3240-0.09a=2920\implies 3240-0.02a=2920\implies -0.02a=-320 \\\\\\ a=\cfrac{-320}{-0.02}\implies \boxed{a=16000}~\hfill \boxed{\stackrel{36000~~ - ~~16000}{20000=b}}

5 0
3 years ago
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