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leva [86]
3 years ago
13

Need help please

Mathematics
1 answer:
svp [43]3 years ago
4 0
-6,6 if it reflects on the left side
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out of 8 marbles, 3 were green. if 37.5% of them were green, what percentage of the marbles were not green
Sidana [21]
100-37.5 = 72.5% of the marbles were not green
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How many miles can you drive with 13.2 gallons of gasoline
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NEED HELP ASAP ILL MAKE BRAINLIEST PLEASE HELP
nordsb [41]

Answer:

33% (B)

Step-by-step explanation

It increased by 150, and 150 is 1/3 of 450, so 33% Hope this helped! Good luck!

6 0
3 years ago
What are the surface area and volume ratios of a cylinder change if the radius and height are multiplied by 5/4 ?
victus00 [196]

Answer:

The ratio of the surface areas and volume is 8((5y+5x) /25xy)

Step-by-step explanation:

This problem bothers on the mensuration of solid shapes.

Let us assume that the radius =x

Radius r=5x/4

And the height =y

Height h= 5y/4

We know that the total surface area of a cylinder is

A total = 2πrh+2πr²

We can factor out 2πr

A total = 2πr(h+r)

The volume of a cylinder is given as

v= πr²h

The surface area and volume ratios

Can be expressed as

2πr(h+r)/πr²h= 2(h+r)/rh

= (2h+2r)/rh= 2h/rh + 2r/rh

= 2/r + 2/h

= 2(1/r + 1/h)

Substituting our value of x and y

For radius and height we have

= 2(1/5x/4 + 1/5y/4)

=2(4/5x + 4/5y)

=2*4(1/5x + 1/5y)

= 8 (5y+5x/25xy)

6 0
3 years ago
Read 2 more answers
$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

4 0
4 years ago
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