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joja [24]
3 years ago
7

Cesar wants to buy grass seed to cover his whole lawn, except for the pool. The pool is 7 1 2 m by 3 1 4 m. Find the area the gr

ass seed needs to cover.
Mathematics
1 answer:
timofeeve [1]3 years ago
4 0

226'394 is the area of grass need to grow

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Are these right?!?! Cannot fail
jolli1 [7]
The first one is correct but I don’t know about the second one
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3 years ago
4
yaroslaw [1]

Answer:

The expression that represents the given sequence is 5+6(n-1). Option C (not labeled).

Explanation:

<u>Arithmetic Sequences</u>

In an arithmetic sequence, each term can be obtained by adding or subtracting a fixed number to the previous term. That fixed number is called the common difference.

We are given the following sequence:

5, 11, 17, 23, 29, ...

Each term is located in a position starting from n=1. Let's test each option:

A For n=1 we should have the first term (5). Substituting n=1 into the general equation: 5+6(n+1) = 5+6(1+1) = 5+12 = 17. Since the resulting term is not 5, this option is incorrect.

B For n=1, 6+5(n+1)= 6+5(2)=16. This option is incorrect.

C (not labeled) For n=1, 5+6(n-1)=5+6(1-1)=5+0=5. The first term is correct. Let's test for the second term (n=2):

5+6(2-1)=5+6=11. Correct. For n=3

5+6(3-1)=5+12=17. Correct.

We can see the terms are increasing by 6, and the given sequence is also increasing by 6. Thus, This option is correct.

D For n=1, 6+5 (n-1)=6+0=6. This option is incorrect.

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svet-max [94.6K]
Y=2x−4 is the graph so you graph it
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Answer:

  B.  x = 4

Step-by-step explanation:

The axis of symmetry is the vertical line through the vertex. This parabola has its vertex at (4, -1), so the axis of symmetry is x = 4.

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A)

\bf \qquad \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\\&#10;P=\textit{original amount deposited}\to &\$15000\\&#10;r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{twelve months, thus}&#10;\end{array}\to &12\\&#10;t=years\to &1&#10;\end{cases}&#10;\\\\\\&#10;A=15000\left(1+\frac{0.0495}{12}\right)^{12\cdot 1}

b)

\bf \qquad \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\\&#10;P=\textit{original amount deposited}\to &\$15000\\&#10;r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{365 days, thus}&#10;\end{array}\to &365\\&#10;t=years\to &1&#10;\end{cases}&#10;\\\\\\&#10;A=15000\left(1+\frac{0.0495}{365}\right)^{365\cdot 1}

c)

\bf \qquad \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\\&#10;P=\textit{original amount deposited}\to &\$15000\\&#10;r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{four quarters, thus}&#10;\end{array}\to &4\\&#10;t=years\to &1&#10;\end{cases}&#10;\\\\\\&#10;A=15000\left(1+\frac{0.0495}{4}\right)^{4\cdot 1}
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