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4 years ago

ASAP* can someone pls help me with this? and can you describe it please?

Mathematics

1 answer:

Studentka2010 [4]4 years ago

5 0

Add 165 to 245. That is a total of 410kWh.

245kWh is more than 50% of 410kWh.

So, according to the plan, you will pay NightTimer rate for kWh at night and DayTimer rate during the day.

The work done:

$(165*0.05281)+(245*0.05456)\\8.71365+13.3674=22.08085$

The raw, accurate, answer is that the value to pay is $22.08085.

In real terms, it is closer to your $22.08 option. That is the one I'd take.

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PLZ ANSWER PIC INCLUDED

Alchen [17]

Answer:

28 feet

Step-by-step explanation:

We know that the perimeter of a square is $2l + 2w$ , where l is the length and w is the width.

We also know that the perimeter of a triangle is all of its sides added up.

Since the perimeters will be equal to each other, we can make both of these statements an equation that are equal to each other.

Basically:

$(3x - 3) + (3x - 3) + (x - 3) + (x - 3) = (2x - 1) + (2x - 1) + (2x)$

We can combine like terms to get a less scary equation.

$8x - 12 = 6x - 2$

Our goal is to now isolate x so we can find it's value.

Add 2 to both sides:

$8x - 10 = 6x$

Subtract 6x from both sides:

$2x - 10 = 0$

Add 10 to both sides:

$2x = 10$

Divide both sides by 2:

$x = 5$

Now we know the value of x, however, our work isn't done. They want to know the perimeter, so we need to substitute x in as 5.

Let's do it on the triangle.

$6x-2$ (remember, this is our simplified version of the triangle)

$6(5) - 2\\\\30 - 2\\\\28$

Hope this helped!

4 0

3 years ago

Each investment matures in 3 years. The interest compounds annually.

Sveta_85 [38]

bearing in mind that 4¾ is simply 4.75.

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$600\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=600\left(1+\frac{0.05}{1}\right)^{1\cdot 3}\implies A=600(1.05)^3\implies A=694.575 \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$750\\ r=rate\to 4.75\%\to \frac{4.75}{100}\dotfill &0.0475\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=750\left(1+\frac{0.0475}{1}\right)^{1\cdot 3}\implies A=750(1.0475)^3\implies A\approx 862.032$