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g100num [7]
3 years ago
7

How can graphing be applied to solving systems of nonlinear equations?

Mathematics
1 answer:
aliina [53]3 years ago
8 0
One example on the top of my head is moors law. It can be graphed as an exponential equation and it models a real world situation
You might be interested in
Proof true or false: For all integers a,b,and c,if ab|c then a|c and b|c
olya-2409 [2.1K]

Answer with explanation:

It is given that for three integers , a, b and c, if

              \frac{ab}{c}\rightarrow then, \frac{a}{c} \text{or} \frac{b}{c}

Since , a b is divisible by c , following are the possibilities

1.→ a and b are prime integers .Then , c will be prime number either equal to a or b.

2.→a and b are not prime integers ,then any of the factors of a or b will be equal to c.For example:

 ⇒a=m × n

 b=p × q× c

or,

⇒a=u×v×c

b=s×t

So, whatever the integral values taken by a, and b, if \frac{ab}{c} then either of  \frac{a}{c} \text{or} \frac{b}{c} is true.

4 0
3 years ago
What is the surface area of this composite figure?
faust18 [17]

Answer:

<h2>          294 cm²</h2>

Step-by-step explanation:

6\cdot6+5\cdot(6\cdot7)+6\cdot4+2\cdot(\frac12\cdot6\cdot4)=36+210+24+24=294\,cm^2

4 0
3 years ago
Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw
vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

<em>x</em> ² - 10<em>x</em> + <em>y</em> ² = 0

<em>x</em> ² - 10<em>x</em> + 25 + <em>y </em>² = 25

(<em>x</em> - 5)² + <em>y</em> ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

<em>x</em> = <em>r</em> cos(<em>θ</em>)

<em>y</em> = <em>r</em> sin(<em>θ</em>)

Then

<em>x</em> ² + <em>y</em> ² = 100   →   <em>r </em>² = 100   →   <em>r</em> = 10

<em>x</em> ² - 10<em>x</em> + <em>y</em> ² = 0   →   <em>r </em>² - 10 <em>r</em> cos(<em>θ</em>) = 0   →   <em>r</em> = 10 cos(<em>θ</em>)

<em>y</em> = 5   →   <em>r</em> sin(<em>θ</em>) = 5   →   <em>r</em> = 5 csc(<em>θ</em>)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to <em>y</em> = 5 at the point (0, 5).

Split up the region at 3 angles <em>θ</em>₁, <em>θ</em>₂, and <em>θ</em>₃, which denote the angles <em>θ</em> at which the curves intersect. They are

<em>θ</em>₁ = 0 … … … by solving 10 = 10 cos(<em>θ</em>)

<em>θ</em>₂ = <em>π</em>/6 … … by solving 10 = 5 csc(<em>θ</em>)

<em>θ</em>₃ = 5<em>π</em>/6  … the second solution to 10 = 5 csc(<em>θ</em>)

Then the area of the region is given by a sum of integrals:

\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)

=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta

To compute the integrals, use the following identities:

sin²(<em>θ</em>) = (1 - cos(2<em>θ</em>)) / 2

cos²(<em>θ</em>) = (1 + cos(2<em>θ</em>)) / 2

and recall that

d(cot(<em>θ</em>))/d<em>θ</em> = -csc²(<em>θ</em>)

You should end up with an area of

=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta

=\boxed{25\sqrt3+\dfrac{125\pi}3}

We can verify this geometrically:

• the area of the larger circle is 100<em>π</em>

• the area of the smaller circle is 25<em>π</em>

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line <em>y</em> = 5, has area 100<em>π</em>/3 - 25√3

Hence the area of the region of interest is

100<em>π</em> - 25<em>π</em> - (100<em>π</em>/3 - 25√3) = 125<em>π</em>/3 + 25√3

as expected.

3 0
3 years ago
Solve the system of equations algebraically. Verify your answer using the graph.
nirvana33 [79]
Both of those equations are solved for y.  So if the first one is equal to y and the second one is equal to y, then the transitive property says that the first one is equal to the second one.  We set them equal to each other and solve for x.  4x-5=-3 and 4x = 2.  That means that x = 1/2.  We were already told that y = -3, so the coordinates for the solution to that system are (1/2, -3), B from above.
6 0
3 years ago
Read 2 more answers
Andrea jumped 5 times every 45 minutes. At that rate, how many
elena55 [62]

Answer:

6 times

Step-by-step explanation:

If Andrea jumps 5 times every 45 minutes, she will jump 1 time every 9 minutes. 54 divided by 9=6 times. Hope it helps!

4 0
3 years ago
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