x = 3 + y Eqn(1)
y = -2x + 9 Eqn(2)
Let us solve the system of equations with the substitution method
x - 3 = y (Subtracting 3 from both sides of the Eqn(1))
Replacing y = x - 3 in Eqn (2), we have:
x - 3 = -2x + 9
x = -2x + 9 + 3 (Adding 3 to both sides of the equation)
x + 2x = 9 + 3 (Adding 2x to both sides of the equation)
3x = 12 ( Adding like terms)
x = 12/3 (Dividing by 3 on both sides of the equation)
x = 4
Replacing x=4 in Eqn(1), we have:
4 = 3 + y
4 - 3 = y (Subtracting 3 from both sides of the equation)
y=1
The answers are:
x= 4 and y=1
The equation which represents the given graph g(x) is 2ˣ-1.
<h3>
What is Equation?</h3>
Equations are mathematical statements containing two algebraic expressions on both sides of an 'equal to (=)' sign.
Here, given graph passes through origin.
Put x = 0 in all the given equation and verify which equations gives result as zero.
Put x = 0 in option A
g(0) = 2⁰⁺¹ = 2 ≠ 0
in option B
g(0) = 2⁰-1 = 1 - 1 = 0
Thus, option B g(x) = 2ˣ-1 is the correct expression for the given graph.
Learn more about Equations from:
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The domain is all real numbers, the y intercept is 3 the asymptote is -1
Im not sure about the last two sorry
<span>−8x+7y=25 is already in standard form, altho some people would prefer to re-write it as
</span><span>−8x+7y-25 = 0.
You have shared 4 equations here. Next time, please separate them with commas or semi colons, or type just 1 equation per line, for increased clarity. Thanks.</span>
The terms 9r^2 and 6r^2 are like terms. We add the numbers to the left of the "r^2" to get 9+6 = 15.
So 9r^2+6r^2 becomes 15r^2
The r^2 stays the same. We do not say something like r^2+r^2 = r^4. You might be thinking of r^2 * r^2 = r^(2+2) = r^4
When it comes to adding like terms, we keep the variables the same.
The other pair of like terms are 3r and 2r, so they add to 3r+2r = 5r
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Overall,
(9r^2+3r+6)+(6r^2+2r) = (9r^2+6r^2)+(3r+2r)+6
(9r^2+3r+6)+(6r^2+2r) = 15r^2+5r+6
