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3 years ago

Plz Halppp

2 answers:

4 0

Hello!

You read scientific notation by moving the decimal that amount of spaces next the the 10

1.263 * 10^3

You move the decimal over 3 places

1.263 * 10^3 = 1263

The answer is 1.263 * 10^3

Hope this helps!

Dmitriy789 [7]3 years ago

4 0

Answer:

Scientific notation 1 .263 × 10³.

Step-by-step explanation:

Given : 1,263

To find : Select the correct scientific notation form of this numeral.

Solution : We have given 1,263

Scientific notation form : scientific notation use a number indicate the powers of 10.

We need to put the decimal after one digit.

Example : 6755 = 6.755* 10³

So, We can write it 1,263 as 1 .263 × 10³.

Therefore, scientific notation 1 .263 × 10³.

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The difference of the square of a number and 4 is equal to 3 times that number. Find the positive solution.

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Let the number be 'a':

The difference of the square of a number and 4

a^2 -4

is equal to 3 times that number.

= 3*a

a^2 -4 =3a

a^2 -3a -4=0

(a-4) (a+1)=0

solutions are: 4 and -1

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Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices

Airida [17]

Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

$\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt$

Where P, Q are scalar functions

We want to compute

$\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy$

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths $\large C_1,C_2,C_3,C_4$ which represents the sides of the rectangle.

$\large C_1$ is the line segment from (0,0) to (5,0)

$\large C_2$ is the line segment from (5,0) to (5,1)

$\large C_3$ is the line segment from (5,1) to (0,1)

$\large C_4$ is the line segment from (0,1) to (0,0)

Then

$\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}$

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize $\large C_1$ as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

$\large \displaystyle\int_{C_1}xydx+x^2dy=0$

Now the second integral. We parametrize $\large C_2$ as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

$\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25$

The third integral. We parametrize $\large C_3$ as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

$\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2$

The fourth integral. We parametrize $\large C_4$ as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

$\large \displaystyle\int_{C_4}xydx+x^2dy=0$

$\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2$

Now, let us compute the value using Green's theorem.

According with this theorem

$\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx$

where A is the interior of the rectangle.

so A={(x,y) | 0≤ x≤ 5, 0≤ y≤ 1}

We have

$\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x$

$\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2$

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Agata [3.3K]

Remark
I take it that you want to know the ratio of the radii. If this is not correct, leave a comment below my answer.

You could do this by giving the spheres a definite volume, like 1 and 8 and then solve for r for one of them and then use the other sphere to find it's radius. It is not exactly the best way, and if you are going to to a physics class you want to be doing this using cancellation.

Step One
Set up the Ratio for the volumes.

$\frac{V_1}{V_2} = \frac{1}{8}$

Step Two
Setup the equation for V1/V2 using the definition for a sphere. V = 4/3 pi r^3

$\dfrac{1}{8} = \frac{ \frac{4}{3} \pi( r_1)^3 }{ \frac{4}{3} \pi( r_2)^3 }$

Step Three
Cancel the 4/3 and pi on the top and bottom of the fractions on the right.

You are left with 1/8 = (r1)^3/ (r2)^3

Step Four
Take the cube root of both sides.
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Cube root of (r1)^3 = r1
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Step Five
Answer
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