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1 year ago

5

milton purchases a 5-gallon aquarium for his bedroom.To fill the aquarium with water,he uses a container with a capacity of 1 qu

art.How many times will milton fill and emty the container before the aquarium is full

2 answers:

goblinko [34]1 year ago

8 0

There are 4 quarts in a gallon to find how many times he will use the container to fill the aquarium you need to multiply 4 by 5:

4×5=20

Milton with fill the container 20 times before the container is full

hope this helped

Charra [1.4K]1 year ago

8 0

He must fill the container 20 times

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How can the triangles be proven similar by the SSS similarity theorem? Show that the ratios StartFraction U V Over X Y EndFracti

dangina [55]

Answer:

<u>option 1</u>

Step-by-step explanation:

The question is as shown in the attached figure.

From the figure, we can deduce that:

∠U ≅ ∠X and ∠W ≅ ∠Z and ∠V ≅ ∠Y

So, ΔUWV similar to ΔXZY

But it is required to know " How can the triangles be proven similar by the SSS similarity theorem? "

So, we need to prove the corresponding angles are proportion

With the help of the previous corresponding angles.

So, we need to show that the ratios $\frac{UV}{XY} \ , \ \frac{WU}{ZX} \ and \ \frac{WV}{ZY}$ are equivalent.

So, the answer is <u>option 1</u>

5 0

1 year ago

At the beginning of year 1, paolo invests $500

suter [353]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$500\\
r=rate\to 4\%\to \frac{4}{100}\to &0.04\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annuall, thus once}
\end{array}\to &1\\
t=years\to &t
\end{cases}
\\\\\\
A=500\left(1+\frac{0.04}{1}\right)^{1\cdot t}\implies A=500(1+0.04)^t
\\\\\\
\textit{after 5 years }t=5\qquad A=500(1.04)^5$

the example on your picture uses A(n) and n = years, but is pretty much the same, in this case is t = years.

3 0

1 year ago

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Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val

Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

$A(t) = Pe^{rt}$

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that $P = 330$

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

$A(t) = Pe^{rt}$

$590 = 330e^{8r}$

$e^{8r} = \frac{590}{330}$

$e^{8r} = 1.79$

Applying ln to both sides:

$\ln{e^{8r}} = \ln{1.79}$

$8r = \ln{1.79}$

$r = \frac{\ln{1.79}}{8}$

$r = 0.0726$

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

$A(t) = 330e^{0.0726t}$

$A(10) = 330e^{0.0726*10} = 682$

The value of the account in the year 2009 will be $682.

4 0

2 years ago

A regular decagon has sides that are 8 cm long. What is the area of the figure? Round to the nearest whole number.

blsea [12.9K]

<u>Given</u>:

Given that the regular decagon has sides that are 8 cm long.

We need to determine the area of the regular decagon.

<u>Area of the regular decagon:</u>

The area of the regular decagon can be determined using the formula,

$A=\frac{s^{2} n}{4 \tan \left(\frac{180}{n}\right)}$

where s is the length of the side and n is the number of sides.

Substituting s = 8 and n = 10, we get;

$A=\frac{8^{2} \times 10}{4 \tan \left(\frac{180}{10}\right)}$

Simplifying, we get;

$A=\frac{64 \times 10}{4 (\tan \ 18)}$

$A=\frac{640}{4 (0.325)}$

$A=\frac{640}{1.3}$

$A=642.3$

Rounding off to the nearest whole number, we get;

$A=642 \ cm^2$

Thus, the area of the regular decagon is 642 cm²

Hence, Option B is the correct answer.

5 0

1 year ago

Which of the following shows the extraneous solution to the logarithmic equation? log Subscript 4 Baseline (x) + log Subscript 4

vekshin1

<u>Given:</u>

The given equation is $\log _{4}(x)+\log _{4}(x-3)=\log _{4}(-7 x+21)$

We need to determine the extraneous solution of the equation.

<u>Solving the equation:</u>

To determine the extraneous solution, we shall first solve the given equation.

Applying the log rule $\log _{c}(a)+\log _{c}(b)=\log _{c}(a b)$ , we get;

$\log _{4}(x(x-3))=\log _{4}(-7 x+21)$

Again applying the log rule, if $\log _{b}(f(x))=\log _{b}(g(x))$ then $f(x)=g(x)$

Thus, we have;

$x(x-3)=-7 x+21$

Simplifying the equation, we get;

$x^2-3x=-7 x+21$

$x^2+4x=21$

$x^2+4x-21=0$

Factoring the equation, we get;

$(x-3)(x+7)=0$

Thus, the solutions are $x=3, x=-7$

<u>Extraneous solutions:</u>

The extraneous solutions are the solutions that does not work in the original equation.

Now, to determine the extraneous solution, let us substitute x = 3 and x = -7 in the original equation.

Thus, we get;

$\log _{4}(3)+\log _{4}(3-3)=\log _{4}(-7 \cdot 3+21)$

$\log _{4}(3)+\log _{4}(0)=\log _{4}(0)$

Since, we know that $\log _{a}(0)$ is undefined.

Thus, we get;

Undefined = Undefined

This is false.

Thus, the solution x = 3 does not work in the original equation.

Hence, x = 3 is an extraneous solution.

Similarly, substituting x = -7, in the original equation. Thus, we get;

$\log _{4}(-7)+\log _{4}(-7-3)=\log _{4}(-7(-7)+21)$

$\log _{4}(-7)+\log _{4}(-10)=\log _{4}(49+21)$

$\log _{4}(-7)+\log _{4}(-10)=\log _{4}(70)$

Simplifying, we get;

Undefined = $\log _{4}(70)$

Undefined = 3.06

This is false.

Thus, the solution x = -7 does not work in the original solution.

Hence, x = -7 is an extraneous solution.

Therefore, the extraneous solutions are x = 3 and x = -7

7 0

1 year ago

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