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3 years ago

What is the length of DE in the right triangle below ?

Mathematics

2 answers:

NeTakaya3 years ago

5 0

Answer: The length of DE is (B) 25 units.

Step-by-step explanation: As shown in the given figure, ΔDEF is a right-angles triangle, where ∠DFE = 90°, DF = 15 units and EF = 20 units.

We are to find the length of DE.

ΔDEF is right-angled at ∠DFE.

So, using Pythagoras theorem, we have

$DE^2=DF^2+EF^2\\\\\Rightarrow DE^2=15^2+20^2\\\\\Rightarrow DE^2=225+400\\\\\Rightarrow DE^2=625\\\\\Rightarrow DE^2=25^2\\\\\Rightarrow DE=25.$

Therefore, the length of DE is 25 units.

Thus, (B) is the correct option.

olganol [36]3 years ago

3 0

B.25
sorry if my writing is not clear

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3 years ago

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kati45 [8]

Answer:

$15 = a_1 r^4$ (1)

$1 = a_1 r^5$ (2)

If we divide equations (2) and (1) we got:

$\frac{r^5}{r^4}= \frac{1}{15}$

And then $r= \frac{1}{15}$

And then we can find the value $a_1$ and we got from equation (1)

$a_1 = \frac{15}{r^4} = \frac{15}{(\frac{1}{15})^4} =759375$

And then the general term for the sequence would be given by:

$a_n = 759375 (\frac{1}{15})^n-1 , n=1,2,3,4,...$

And the best option would be:

C) a1=759,375; an=an−1⋅(1/15)

Step-by-step explanation:

the general formula for a geometric sequence is given by:

$a_n = a_1 r^{n-1}$

For this case we know that $a_5 = 15, a_6 = 1$

Then we have the following conditions:

$15 = a_1 r^4$ (1)

$1 = a_1 r^5$ (2)

If we divide equations (2) and (1) we got:

$\frac{r^5}{r^4}= \frac{1}{15}$

And then $r= \frac{1}{15}$

And then we can find the value $a_1$ and we got from equation (1)

$a_1 = \frac{15}{r^4} = \frac{15}{(\frac{1}{15})^4} =759375$

And then the general term for the sequence would be given by:

$a_n = 759375 (\frac{1}{15})^n-1 , n=1,2,3,4,...$

And the best option would be:

C) a1=759,375; an=an−1⋅(1/15)

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3 years ago