All categories

3 years ago

Fill in the missing monomials: a 100x2= (_____)2

Mathematics

2 answers:

zysi [14]3 years ago

6 0

Answer:

The anser is 100

Step-by-step explanation:

When there is a number right next to parenthisis you mulptiply. SO technicly the (100)2 is just 100*2.

nikitadnepr [17]3 years ago

3 0

Answer:

100

Step-by-step explanation: its the same thing but in a different form

You might be interested in

1. The backdrop of the altar in a church is triangular in shape. It has a base of 10.5 meters

FrozenT [24]

Answer:

Area of triangle backdrop:

A = 1/2bh = 1/2(10.5)(8.5) = 44.63 m²

Area of the cloth:

A = 4.5*4 = 18 m²

Uncovered area:

44.63 - 18 = 26.63 m²

Choice A

Area of the cut piece:

A = πr² = 3.14*7.6² = 181.37 cm²

Area of the square piece:

A= a² = 34² = 1156 cm²

Remaining piece is:

1156 - 181.37 = 974.63 cm²

Choice B

8 0

3 years ago

Simplify the expression 13 (9x)

Ratling [72]

13(9x)

13 * 9x = 117x

5 0

3 years ago

Read 2 more answers

There is a store downtown that sells clothes. The owner has his own method of pricing items. A coat costs $20, a hat costs $15,

Lesechka [4]

Answer:

$25

Step-by-step explanation:

well you have to identify what the method is first.

He is basically pricing things by the count of 5. So, you can assume that it would cost $25.

Idk, this is a weird problem

7 0

3 years ago

360,000 = X% of 4000000

shutvik [7]

Answer:9%

360,000 / 4000000=.09

.09x100= 9%

6 0

3 years ago

Read 2 more answers

Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) t

pashok25 [27]

Direct computation:

Parameterize the top part of the circle $x^2+y^2=1$ by

$\vec r(t)=(x(t),y(t))=(\cos t,\sin t)$

with $0\le t\le\pi$ , and the line segment by

$\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)$

with $0\le t\le1$ . Then

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)$

$=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt$

$=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}$

Using the fundamental theorem of calculus:

The integral can be written as

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}$

If there happens to be a scalar function $f$ such that $\vec F=\nabla f$ , then $\vec F$ is conservative and the integral is path-independent, so we only need to worry about the value of $f$ at the path's endpoints.