I need help..............

Mathematics

1 answer:

vovangra [49]3 years ago

7 0

Answer:

$a_n=-3+7(n-1)$

Step-by-step explanation:

The recursive rule tells you the initial term of the sequence is a1 = -3, and the common difference is d=7. (7 is the value added to one term to get the next term.)

Putting these values into the formula for the explicit rule gives ...

an = a1 +d(n -1)

an = -3 + 7(n -1)

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Estimate 154 divide 7<br>

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2 years ago

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PLZ HELP, AND PLZ EXPLAIN

shutvik [7]

Answer:

Step-by-step explanation:

To make it easy let's start by organizing our information :

AC=12 AND BD=8
ABCD is a rhombus
K and L are the midpoints of sides AD and CD
we notice that the rhombus ABCD is divided into four right triangles

What do you think of when you hear a right triangle ?

The pythagorian theorem !

AC and BD are khown so let's focus on them .

If we concentrated we can notice that AB and BD are cossing each other in the midpoints . why ?

Simply because they are the diagonals of a rhombus .

ow let's apply the pythagorian theorem :

(AC/2)² + (BD/2)² = BC²
6²+4²=52
BC²= 52⇒ $\sqrt{52}$ =BC

Now we khow that : AB=BC=CD=AD= $\sqrt{52}$

This isn't enough . Let's try to figure out a way to calculate the length of KL wich is the base of the triangle

KL is parallel to AC
k is the midpoint of AD and L of DC

I smell something . yes! Thales theorem

KL/AC=DL/DC=DK/AD WE4LL TAKE OLY ONE

KL/12= $\sqrt{52}$ /2* $\sqrt{52}$
KL/12=1/2⇒ KL=6

Now we have the length of the base kl

Now the big boss the height :

notice that you khow the length of KL
BD crosses kl from its midpoint and DL = $\sqrt{52}$ /2

What I want to do is to apply the pythgorian thaorem to khow the lenght of that small part that is not a part of the height of the triangle . I will call it D