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4 years ago

7

P is the midpoint of DE¯¯¯¯¯DE¯ , DP=3x+2DP=3x+2, and DE=10x−12DE=10x−12. What is the length of DP¯¯¯¯¯DP¯ ?

2 answers:

miss Akunina [59]4 years ago

5 0

3x+2 mulitiply by 2. you will get 6x+4 now put 10x-12 like so 6x+4=10x-12. then use like terms and subtract 6x from 10x. you will get 4x-12=4. then add 12 to cancel out then divide 16 by 4 and then x=4 then plug in 4 to replace x in 3x+2.
The answer is 14

Ne4ueva [31]4 years ago

3 0

Answer:

14 units

Step-by-step explanation:

done the topic

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3 years ago

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Step-by-step explanation:

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3 years ago

In an arithmetic sequence, the nth term an is given by the formula an=a1+(n−1)d, where a1 is the first term and d is the commo

Dmitry_Shevchenko [17]

Answer:

$a_{10} = \frac{10}{65536}$

Step-by-step explanation:

The first step to solving this problem is verifying if this sequence is an arithmetic sequence or a geometric sequence.

This sequence is arithmetic if:

$a_{3} - a_{2} = a_{2} - a_{1}$

We have that:

$a_{3} = 40, a_{2} = 10, a_{3} = \frac{5}{2}$

$a_{3} - a_{2} = a_{2} - a_{1}$

$\frac{5}{2} - 10 = 10 - 40$

$\frac{-15}{2} \neq -30$

This is not an arithmetic sequence.

This sequence is geometric if:

$\frac{a_{3}}{a_{2}} = \frac{a_{2}}{a_{1}}$

$\frac{\frac{5}[2}}{10} = \frac{10}{40}$

$\frac{5}{20} = \frac{1}{4}$

$\frac{1}{4} = \frac{1}{4}$

This is a geometric sequence, in which:

The first term is 40, so $a_{1} = 40$

The common ratio is $\frac{1}{4}$ , so $r = \frac{1}{4}$ .

We have that:

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

The 10th term is $a_{10}$ . So:

$a_{10} = a_{1}*r^{9}$

$a_{10} = 40*(\frac{1}{4})^{9}$

$a_{10} = \frac{40}{262144}$

Simplifying by 4, we have:

$a_{10} = \frac{10}{65536}$

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4 years ago

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Answer:

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Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Given the two functions, which statement is true?

tia_tia [17]

Answer:

Option D

Step-by-step explanation:

f(x) = $\text{log}_{15}x$

Transformed form of the function 'f' is 'g'.

g(x) = $\frac{1}{2}\text{log}_{15}(x+4)$

Property of vertical stretch or compression of a function,

k(x) = x

Transformed function → m(x) = kx

Here, k = scale factor

1). If k > 1, function is vertically stretched.

2). If 0 < k < 1, function is vertically compressed.

From the given functions, k = $\frac{1}{2}$

Since, k is between 0 and $\frac{1}{2}$ , function f(x) is vertically compressed by a scale factor $\frac{1}{2}$ .

g(x) = f(x + 4) represents a shift of function 'f' by 4 units left.

g(x) = f(x - 4) represents a shift of function 'f' by 4 units right.

g(x) = $\frac{1}{2}\text{log}_{15}(x+4)$

Therefore, function f(x) has been shifted by 4 units left to form image function g(x).

Option D is the answer.

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3 years ago