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3 years ago

15

***Question is on the picture

1 answer:

RUDIKE [14]3 years ago

5 0

Answer:

M(4, -3)

Step-by-step explanation:

Reflecting over the x-axis is simple:

If the x is positive, the reflection would be negative and vice versa.

In this case, the <em>y </em>isn't affected since it's only reflecting over the x-axis.

So, M is at the coordinates (4, -3)

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Answer:

The answer is option 1.

Step-by-step explanation:

You have to apply Gradient formula :

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$let \: (x2 \: , \: y2) \: be \: (0 \: , \: 4)$

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$m = 2$

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Write an explicit rule in function form for the sequence represented by the given terms.

hram777 [196]

Answer:

The general relation about a geoemtric sequence is

$a_{n}=a_{1}r^{n-1}$

For the first sequence, we have

$3=a_{1}r^{5-1}$

$147=a_{1}r^{7-1}$

Which is a system of equations. We can isolate a variable in the first equation to replace that expression into the second equation.

$a_{1}=\frac{3}{r^{4} }$

$147=\frac{3}{r^{4} }r^{6}$

$r^{2}=\frac{147}{3}\\ r=\sqrt[2]{49}\\r=7$

Now, we replace this value to find the other one

$3=a_{1}(7)^{4}\\ a_{1}=\frac{3}{2401}$

Therefore, the explicit rule function is

$a_{n}=\frac{3}{2401} \times (7)^{n-1}$

Now, we use the same process for the second sequence.

$10=a_{1}r^{3-1} \\a_{1}=\frac{10}{r^{2} }$

The second equation is

$1440=a_{1}r^{5-1}\\a_{1}=\frac{1440}{r^{4} }$

Now, we solve the following expression

$\frac{10}{r^{2} }=\frac{1440}{r^{4} }$

We solve for $r$

$\frac{r^{4} }{r^{2} }=\frac{1440}{10}\\r^{2}=144\\ r=\sqrt{144} \\r=12$

Then

$a_{1}=\frac{10}{(12)^{2} } =\frac{10}{144}=\frac{5}{72}$

Therefore, the function that models the second sequence is

$a_{n}=\frac{5}{72} \times (12)^{n-1}$

Notice that $a_{n}$ is the dependent variable and $n$ is the independent variable.

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3 years ago