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2 years ago

How do you solve this using system of equations y = 3x -1 y = -x +3

Mathematics

2 answers:

Allisa [31]2 years ago

4 0

Answer:

Step-by-step explanation:

The first equation's y value can be replaced with -x + 3, then solve it:

3x - 1 = -x + 3

3x + x = 3 + 1

4x = 4

x = 4 divided by 4

x = 1

nikklg [1K]2 years ago

4 0

Answer:

Step-by-step explanation:

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Five line segments coincide at a point as shown.

Nikitich [7]

Answer:

720º

Step-by-step explanation:

We can use vertical angles to figure this out. We can see that each of the unmarked angles in the triangles is a vertical angle to an empty space. A full circle has the measure of 360º. We need half of that which is 180º.

The sum of the angle measures of a triangle is 180º. We have five triangles so the total angle meaure is 900º.

However, we need to subtract the 180º to get 720º.

8 0

2 years ago

Suppose that surface σ is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩, 0≤u≤7 and 0≤v≤2π3 and f(x,y,z)=x2+y2+z2. Set up the sur

Bad White [126]

Looks like you have most of the details already, but you're missing one crucial piece.

$\sigma$ is parameterized by

$\vec r(u,v)=\langle u\cos3v,u\sin3v,v\rangle$

for $0\le u\le7$ and $0\le v\le\frac{2\pi}3$ , and a normal vector to this surface is

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left\langle\sin3v,-\cos3v,3u\right\rangle$

with norm

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{\sin^23v+(-\cos3v)^2+(3u)^2}=\sqrt{9u^2+1}$

So the integral of $f(x,y,z)=x^2+y^2+z^2$ is

$\displaystyle\iint_\sigma f(x,y,z)\,\mathrm dA=\boxed{\int_0^{2\pi/3}\int_0^7(u^2+v^2)\sqrt{9u^2+1}\,\mathrm du\,\mathrm dv}$

6 0

3 years ago

Complete the expressions.

solong [7]

Complete question :

Complete the expressions Write each answer as a number, a variable, or the product of a number and a variable 7(9r + 2) = 7 . 9r + 7 . ___ = ___ +

Answer:

2 ; 63r ; 14

Step-by-step explanation:

Given :

7(9r + 2)

Opening the bracket :

7*9r + 7*2 - - - - (1)

Taking the product

63r + 14 - - - - (2)

Now filling the blanks :

First blank corresponds to 2 (from (1))

Second blank corresponds to 63r (from (2))

Third blank should be 14

8 0

3 years ago

Create a table of data for two different linear functions. The table should use the same values of x for both functions. Based o

user100 [1]

Answer:

Yes they will intersect

Function 1= F(X)=2X+5

Function 2=H(X)=3X+2

INTERSECT=(3,11)

Step-by-step explanation:

First of all, we create 2 LINEAR function, i created the function f(x)=2x+5 and the function h(x)=3x+2, both are linear(without a quadratic term). Then

you replace the x for a number:

Table 1 (F(X)=2X+5) Table 2 (H(X)=3X+2)

X=1----->Y=2+5=7 X=1------>Y=3·1+2=5

X=2---->Y=2·2+5=9 X=2----->Y=3·2+2=8

X=3---->Y=3·3+5=11 X=3----->Y=3·3+2=11

With both tables of data we can see that in the X=3/Y=11 point this two linear functions will intersect so the answer is that the two functions will intersect at (3,11)----->(X,Y)

4 0

3 years ago

Write a polynomial of degree 5 with zero x=0,i square root 7, -2i

professor190 [17]

Answer:

$P(x)=x^5+11x^3+28x$

Step-by-step explanation:

<u>Roots of a polynomial</u>

If we know the roots of a polynomial, say x1,x2,x3,...,xn, we can construct the polynomial using the formula

$P(x)=a(x-x_1)(x-x_2)(x-x_3)...(x-x_n)$

Where a is an arbitrary constant.

We know three of the roots of the degree-5 polynomial are:

$x_1=0;\ x_2=\sqrt{7}\boldsymbol{i}:\ x_3=-2\boldsymbol{i}$

We can complete the two remaining roots by knowing the complex roots in a polynomial with real coefficients, always come paired with their conjugates. This means that the fourth and fifth roots are:

$x_4=-\sqrt{7}\boldsymbol{i}:\ x_3=+2\boldsymbol{i}$