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2 years ago

You have to find the domain and range

Mathematics

1 answer:

Dahasolnce [82]2 years ago

3 0

Answer:

First off, we look for which circles are open or closed.

We start with an open interval since the circle on the left is open and end with a closed interval since the circle on the right is closed.

Domain is all x values, Range is all y values

The graph shows the continous function going from -3 to 1 on the x axis.

According to the circles, this means our domain will be (-3,1].

Now, the range doesn't care about if its closed or not. So we can say the graph is on the y axis from -4 and 0. This means the range is -4<y<0

I used different notations for both just incase you need to represent your answer differently :)

-3<x<1 & (-3,1] . Range is [-4,0]. 0>y>-4 looks correct as-well.

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3x + 2 = 65 * What is x

KiRa [710]

3x = 65 - 2
3x = 63
x = 21

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2 years ago

Suppose that F is an inverse square force field below, where c is a constant.

Andre45 [30]

Answer:

Forces in our Universe

Step-by-step explanation:

First of all we have,

$F(r) = \frac{cr}{|r|^{3} }$

and,

$r = xi+yj+zk$

We need to define a function that allows us to find said change based on r, so one of the functions that shows that change is,

$f(r) = - c /|r|$

That is,

$\nabla f = F$

For this case F is a conservative field and the line integral is independent of the path. Thus, defining $P_{1} = (x_{1}, y_{1}, z_{1})$ and $P_{2} = (x_{2}, y_{2}, z_{2})$ . So the amount of work on the movement of the object from P1 to P2 is,

$W=\int\limits_c {F} \, dr$

$W= f(P_{1}-P_{2})$

$W= \frac{c}{(x_{2}^2+ y_{2}^2+ z_{2}^2)^{1/2} } -\frac{c}{(x_{1}^2+ y_{1}^2+ z_{1}^2)^{1/2}}$

$W= c(\frac{1}{d1}-\frac{1}{d2} )$

2) The gravitational force field is given by,

$F(r) =-\frac{mMGr}{ |r|^3}$

The maximum distance from the earth to the sun is $1.52*10 ^ 8$ km and the minimum distance is $1.47*10 ^ 8$ km. The mass values of the bodies are given by m = $5.97*10 ^ {24}$ kg, M = $1.99 *19 ^ {30}$ kg and the constant G is $6.67 * 10 ^{ -11 }$ $\frac{Nm ^ 2}{kg^2}$

In this way we raise the problem like this,

$c= -mMG$

$W= -mMG (\frac{1}{1.52*10 ^ 8} -\frac{1}{1.47*10 ^ 8} )$

$W= -(5.95*10^{24})(1.99*10^{30})(6.67*10^{-11})(-2.2377*10^{-10})$

$W \approx 1.77*10^{35}}J$

5 0

2 years ago

Help pls and thxxxxxxx

tangare [24]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A box has a volume of 308 cm cubed. Its height is 4 cm greater than its length. Its length is 3 cm greater than its width. What

kifflom [539]

Answer:

4 cm.

Step-by-step explanation:

Given:

A box has a volume of 308 cm cubed.

Its height is 4 cm greater than its length.

Its length is 3 cm greater than its width.

Question asked:

What is the width of the box?

Solution:

Let width = $x\ cm$

Length = $x+3\ cm$

Height = $x+3+4=x+7\ cm$

As we know:

$Volume\ of\ box=length\times width\times height$

$308=(x+3)\times x\times(x+7)\\ \\ 308=(x^{2} +3x)(x+7)\\ \\ 308=x^{2} (x+7)+ 3x(x+7)\\ \\ 308=x^{3} +7x^{2} +3x^{2} +21x\\ \\ 308=x^{3} +10x^{2} +21x\\ \\ Subtract\ 308\ from\ both\ sides\\ \\ x^{3} +10x^{2} +21x-308=0\\ \\ factor\ left\ side\ of\ equation\\ \\ (x-4)(x^{2} +14x+77)=0\\ \\ Set\ factors\ equal\ to\ 0.\\ \\ x-4=0 \ or\ x^{2} +14x+77=0\\ \\ \\$

As width can never be in negative, $x= 4\ cm$

By substituting the value:-

Width = $x\ cm$ = 4 cm

Thus, width of the box is 4 cm.

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2 years ago

Evaluting Algebraic Expressions

Rzqust [24]

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3(6) + 4
18 + 4
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2. N/2 - 1
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3 - 1
2

3. 15 - n
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4. n² - 5
6² - 5
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5. 9(N-2)
9(6-2)
9(4)
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6. 4n / 3+1
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7. 2(18-n)
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Hope this helped!! If you liked my answer, please mark me as brainliest. Thank you! :9

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3 years ago