Brian has started a savings account. The balance, g(x), in his account can be represented by the function g(x) = 25x + 350, where x is the number of weeks he deposits money. Suppose 25 changed to 35. What does this mean in the context of the problem

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Brian has started a savings account. The balance, g(x), in his account can be represented by the function g(x) = 25x + 350, where x is the number of weeks he deposits money. Suppose 25 changed to 35. What does this mean in the context of the problem

Mathematics
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25x is variable amount
answer choices changing 25 to 35 means his initial deposit increased changing 25 to 35 means his initial deposit decreased changing 25 to 35 means the amount he deposits each week increased changing 25 to 35 means the amount he deposits each week decreased
C

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\(\large \begin{array}{llll} g(x) = &{\color{brown}{ 25 }} x + &350\\ &{\color{brown}{ slope}}&y-intercept \end{array}\qquad slope=\textit{rate of change}\)
he deposits in his acccount every "x" amount of weeks originally it'd be every 25x weeks if 25 changes to 35, then it'd be 35x so he'd be depositing every 35x weeks
the hmm wording is a bit misleading 25x means... every "x" week he deposits 25 bucks so say on week 1, x =1 and his deposit is 25(1) week 2, x = 2 and his deposit is 25(2) and so on so if 25 changes to 35 then on week 1, he'd be depositing 35(1) and on week 2, 35(2) and so on
Write the equation of a line with a slope of –1 and a y-intercept of –6
is it y =-x-6
\(\bf \begin{array}{llll} f(x) = &{\color{brown}{ -1 }} x + &-6\\ &{\color{brown}{ slope}}&y-intercept \end{array}\) graph it away
yeap
Write the equation of the line that passes through (−3, 5) and (2, 10) in slope-intercept form
i think it is y=-5x-10
\(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -3}}\quad ,&{\color{blue}{ 5}})\quad % (c,d) &({\color{red}{ 2}}\quad ,&{\color{blue}{ 10}}) \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}} \\ \quad \\ % point-slope intercept y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\ \qquad \uparrow\\ \textit{point-slope form} \)
same as the one before... you may want to post anew, more eyes btw
slope is -5
y-int is -10
i did that and i got y=5x+8
hmmm got a typo darn it \(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -3}}\quad ,&{\color{blue}{ 5}})\quad % (c,d) &({\color{red}{ 2}}\quad ,&{\color{blue}{ 10}}) \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ 10}}-{\color{blue}{ 5}}}{{\color{red}{ 2}}-{\color{red}{ (-3)}}}\implies \cfrac{5}{2+3}\implies \cfrac{\cancel{5}}{\cancel{5}}\implies 1 \\ \quad \\ % point-slope intercept y-{\color{blue}{ 5}}={\color{green}{ 1}}(x-{\color{red}{ (-3)}})\qquad \textit{plug in the values and solve for "y"}\\ \qquad \uparrow\\ \textit{point-slope form}\)

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