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3 years ago

ILL VOTE YOU BRAINLIEST

Mathematics

2 answers:

tangare [24]3 years ago

6 0

Answer:

Step-by-step explanation:

Natali [406]3 years ago

3 0

The answer is : B. -9

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Bruce is going grocery shopping. he can go to the deli, the bakery, and the dairy section. How many possibilities are there for

vodomira [7]

Answer:

1. deli, bakery, dairy;

2. deli, dairy, bakery;

3. bakery, dairy, deli;

4. bakery, deli, dairy;

5. dairy, bakery, deli;

6. dairy, deli, bakery

Step-by-step explanation:

6 possibilities

8 0

3 years ago

What expression is equivalent to the area of metal sheet required to make this square-shaped traffic sign?

AveGali [126]

Answer: $\text{Area of the square shaped traffic sign }=16x^2+9+24x$

Step-by-step explanation:

Given: The side of the square shaped traffic sign = 4x+3

We know that the area of square is given by:-

$Area= side^2$

Therefore, the area of the side of the square shaped field is given by:-

$Area=(4x+3)^2$

We know that , $(a+b)^2=a^2+b^2+2ab$

Therefore,

$(4x+3)^2=(4x)^2+(3)^2+2(4x)(3)\\=16x^2+9+24x$

Hence, $\text{Area of the square shaped traffic sign }=16x^2+9+24x$

3 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.