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3 years ago

Find the ninth partial sum of infinite sumnation 3i-4

1 answer:

3 0

$\bf \sum\limits_{i=1}^9\ 3i-4\implies \sum\limits_{i=1}^9\ 3i-\sum\limits_{i=1}^9\ 4\implies 3\sum\limits_{i=1}^9\ i-\sum\limits_{i=1}^9\ 4
\\\\\\
3\cdot \cfrac{9(9+1)}{2}-(9\cdot 4)$

and surely you know how much that is.

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In triangle ΔABC, ∠C is a right angle and CD is the height to

Zina [86]

Answer:

$m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha$

Step-by-step explanation:

The triangles are drawn below.

CD is perpendicular to AB as CD is height to AB.

Therefore, angles $m\angle CDB=m\angle CDA=90$ °

So, triangles ΔCBD and ΔCAD are right angled triangles.

Now, from the right angled triangle ΔABC,

$m\angle A+m\angle B =90\\\alpha+m\angle B=90\\m\angle B=90-\alpha$

From ΔCBD,

$m\angle CBD$ is same as $m\angle B$ .

So, $m\angle CBD=90-\alpha$

$m\angle BCD+m\angle BDC =90\\m\angle BCD+90-\alpha=90\\m\angle BCD=\alpha$

Now, from ΔCAD,

$m\angle CAD$ is same as $m\angle A$

So, $m\angle CAD=\alpha$

$m\angle CAD+m\angle ACD =90\\\alpha+m\angle ACD=90\\m\angle ACD=90-\alpha$

Hence, the unknown angles of both the triangles are:

$m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha$

5 0

3 years ago

The sun produces 3.8x10^27 joules of energy per second. How much energy is produced in a year? (Note: a year is approximately 31

Slav-nsk [51]

About 1.204972095x10^20 joules in a year

3 0

3 years ago

Is this expression fully simplified? Explain<br> 4x2 + 10 - x - 6x3

tankabanditka [31]

Answer:

-x

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Help me.................

hodyreva [135]

I think it would be letter D

4 0

3 years ago

Read 2 more answers

Let f(x) = -2x + 4 and g(x) = 5x - 7. Find f(x) - g(x).

balandron [24]

Answer:

If you are just looking for a simplified answer then the answer is -7x - 3.

4 0

3 years ago