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2 years ago

What is the coefficient of the term x5y7 in the expansion of (x - y)12 ?

Mathematics

1 answer:

lozanna [386]2 years ago

6 0

Answer:

12 and -12

Step-by-step explanation:

12x-12y is the expansion using the distributive property, so the coefficient of x is 12 and the coefficient of y is -12

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Please help me answer the question.ghvjhbikb

rjkz [21]

Answer:

240units^2

Step-by-step explanation:

The formula for area on a triangle is 1/2 x b x h, so you do 4 x 10 x 1/2, giving you 20, there are two of these triangles so you double it to 40.

Then for the rectangle below you do l x w, or 10 x 20 giving 200.

You then add it all together, 200+20+20=240 units squared.

Hope this helps!

6 0

3 years ago

A rectangular banner has a length that is two feet shorter than twice its height, ℎ. If the expression ℎ(2ℎ−2) represents the ar

densk [106]

Answer:

h = 6

Step-by-step explanation:

Given he area of the banner expressed as;

A = ℎ(2ℎ−2)

h is the height of the banner

A is the area = 60

Substitute

60 = ℎ(2ℎ−2)

60 = 2h² - 2h

30 = h² - h

h²-h-30 = 0

Factorize;

h²-6h+5h-30 = 0

h(h-6)+5(h-6) = 0

(h-6)(h+5) = 0

h - 6 = 0 and h+5 = 0

h = 6 and -5

Since the height cannot be negative;

h = 6

Hence he height of the banner is 6

3 0

3 years ago

Jeremy has a total of $250. If he spends 30% of his money, how much will he have left?

vagabundo [1.1K]

If you look at the picture, you set it up as part over whole is equal to percent over 100. then you just multiply across.

5 0

3 years ago

Which graph represents the solution set of this inequality?<br> 119 +5 < 49<br> Choose 1 answer:

laiz [17]

Answer:

to graph an inequality you need a variable. your question does not have a variable. please rewrite the question.

6 0

2 years ago

Find the surface area of the surface given by the portion of the paraboloid z=3x2+3y2 that lies inside the cylinder x2+y2=4. (hi

natta225 [31]

Parameterize the part of the paraboloid within the cylinder - I'll call it $S$ - by

$\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\left\langle u\cos v,u\sin v,3u^2\right\rangle$

with $0\le u\le2$ and $0\le v\le2\pi$ . The region's area is given by the surface integral

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=2}\int_{v=0}^{v=2\pi}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du\,\mathrm dv$
$=\displaystyle2\pi\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du$

Take $w=1+36u^2$ so that $\mathrm dw=72u\,\mathrm du$ , and the integral becomes

$=\displaystyle\frac{2\pi}{72}\int_{w=1}^{w=145}\sqrt w\,\mathrm dw$
$=\displaystyle\frac\pi{36}\frac23w^{3/2}\bigg|_{w=1}^{w=145}$
$=\dfrac\pi{54}(145^{3/2}-1)\approx101.522$

7 0

2 years ago