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2 years ago

The first three terms of an arithmetic sequence are as follows. -7 , -10 ,-13 Find the next two terms of this sequence.

Mathematics

1 answer:

kotykmax [81]2 years ago

7 0

Step-by-step explanation:

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jolli1 [7]

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2 years ago

Read 2 more answers

Hey, I need help with questions 1 and 2, if anyone can help, please? thanks!

Eva8 [605]

The correct works are:

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$ .
$\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h$

<h3>Function Notation</h3>

The function is given as:

$Blue(s) = 2s^2 + 3$

The interpretation when Steven is asked to calculate Blue(s + h) is that:

Steven is asked to find the output of the function Blue, when the input is s + h

So, we have:

$Blue(s + h) = 2(s + h)^2 + 3$

Evaluate the exponent

$Blue(s + h) = 2(s^2 + 2sh + h^2) + 3$

Expand the bracket

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$

So, the correct work is:

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$

<h3>Simplifying Difference Quotient</h3>

In (a), we have:

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$

$Blue(s) = 2s^2 + 3$

The difference quotient is represented as:

$\frac{f(x + h) - f(x)}{h}$

So, we have:

$\frac{Blue(s + h) - Blue(s)}{h} = \frac{2s^2 + 4sh + 2h^2 + 3 - 2s^2 - 3}{h}$

Evaluate the like terms

$\frac{Blue(s + h) - Blue(s)}{h} = \frac{4sh + 2h^2}{h}$

Evaluate the quotient

$\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h$

Hence, the correct work is:

$\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h$

Read more about function notations at:

brainly.com/question/13136492

4 0

1 year ago

HELP PLEASE!!!!

kumpel [21]

Answer:

The rational numbers are $\frac{11}{3}, 6.25, 0.01045,\sqrt{\frac{16}{81}},0.\bar{42}$ and the irrational functions are $\sqrt{48},\sqrt{\frac{3}{16}}$ .

Step-by-step explanation:

A rational number can be expressed in the form of $\frac{p}{q}$ , where p and q are integers and q is not equal to 0. For example $2,3.5,\frac{2}{5},...$ .

An irrational function can not be expressed in the form of $\frac{p}{q}$ , where p and q are integers and q is not equal to 0. For example $\sqrt{2},\sqrt{3},\sqrt{5},...$ .