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3 years ago

I was wondering if how you would plot a point using this information, also how you would answer this question. Thanks!

Mathematics

1 answer:

lukranit [14]3 years ago

4 0

for the y value you would go up the top line by 2 in problem 3 for the x value you would go to the left by 4 and follow the same steps but with different numbers for problem 4

Have a great day

Hope that helps :)

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Step-by-step explanation:

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3 years ago

Write an equation in slope intercept form of the line that passes through the given point and is parallel to the given equation.

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Slope=-3
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3 years ago

Veronica saved up her tutoring money to buy video games. She buys 2 games for $30.00 each and a game controller for $19.59. Vero

tatuchka [14]

Answer: See explanation

Step-by-step explanation:

Since Veronica buys 2 games for $30.00 each and a game controller for $19.59. The total cost to be paid will be:

= (2 × $30.00) + $19.59

= $60.00 + $19.59

= $79.59

Since Veronica lives in San Jose, California, where sales tax is 9.25%, the sales tax will be:

= 9.25% × $79.59

= 0.0925 × $79.59

= $7.36

Total amount paid = $79.59 + $7.36 = $86.95

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3 years ago

$20000 is invested in an account that earned 6% p.A. Compounding yearly for 3 years. The interest rate then went up to 8% p.A. F

GuDViN [60]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the first 3 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$20000\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases}$

$\bf A=20000\left(1+\frac{0.06}{1}\right)^{1\cdot 3}\implies A=20000(1.06)^3\implies \boxed{A=2382.032} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the next 4 years}}$

$\bf A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2382.032\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}$

$\bf A=2382.032\left(1+\frac{0.08}{1}\right)^{1\cdot 4}\implies A=2382.032(1.08)^4\implies \boxed{A\approx 3240.73} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{amount for this period}}{2382.032+3240.73}\implies 5622.762$

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4 years ago

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