All categories

2 years ago

Can someone help me please

Mathematics

1 answer:

Anna007 [38]2 years ago

6 0

Answer:

The Area of the shaded region :

(x²)(4x-1) - (5x)(x+7) =

4x³ - x² - 5x² - 35x =

4x³ - 6x² - 35x

You might be interested in

Find value of cos75 degrees

AfilCa [17]

Answer:

0.258819045...

Step-by-step explanation:

just plug it into a calculator

7 0

2 years ago

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

5 0

3 years ago

I dont know what a complementary angle is, can someone help me with this quistion please?

Allisa [31]

Answer:

A, complementary angles sum up to 90°

8 0

3 years ago

30 POINTS!! ASAP! PLS SHOW WORK

sp2606 [1]

Answer:

QII

Step-by-step explanation:

We can use the acronym: ASTC or All Students Take Calculus.

The first letter of each word tells us details within a certain quadrant.

All trig functions in QI will be positive.

Only sine (and cosecant) will be positive in QII.

Only tangent (and cotangent) will be positive in QIII.

And only cosine (or secant) will be positive in QIV.

We know that tan(θ)<0. In other words, it's negative.

So, it our angle θ <em>cannot</em> be in QI or QIII.

We also know that cos(θ)<0.

So, this removes QIV, since in QIV, cosine is positive.

Therefore, the only choices that remains is QII.

In QII, sine is positive, tangent is negative, and cosine is negative.

This fits all our conditions, so θ is in QII.

And we're done!

3 0

3 years ago

Please help me im very depressed and im struggling a lot on my math and i really need some answers

expeople1 [14]

pls dont cheat on quiz: ask for extra practice from teacher and ask questions if you don't get material

4 0

3 years ago

Read 2 more answers