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2 years ago

Find the ordered triple of these equations. x + 2y + 2z = 02x + y + z = 3x - 3y - z = 1

1 answer:

3 0

{x+2y+2z=0
{2x+y+z=3
{x-3y-z=1

{x+2y+2z=0
-
{4x+2y+2z=6
{-3x=-6 → x=2

{-3y-z=-1
+
{y+z=-1
{-2y=-2 → y=1

{2+2+2z=0 → 2z=-4 → z=-2
Final triple is
(2; 1; -2)

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Please help!! What is the measure of angels DEF and angles DEG?

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2 years ago

A frustum is made by removing a small cone from a similar large cone in the diagram shown the height of a small cone is a quarte

lesantik [10]

Answer:

Volume of the frustum = ⅓πh(4R² - r²)

Step-by-step explanation:

We are to determine the volume of the frustum.

Find attached the diagram obtained from the given information.

Let height of small cone = h

height of the large cone = H

The height of a small cone is a quarter of the height of the large cone:

h = ¼×H

H = 4h

Volume of the frustum = volume of the large cone - volume of small cone

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6 0

2 years ago

A line passes through the point (–2, 7) and has a slope of –5.

kakasveta [241]

Answer:

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Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Let C = C1 + C2 where C1 is the quarter circle x^2+y^2=4, z=0,from (0,2,0) to (2,0,0), and where C2 is the line segment from (2,

trapecia [35]

Not much can be done without knowing what $\mathbf F(x,y,z)$ is, but at the least we can set up the integral.

First parameterize the pieces of the contour:

$C_1:\mathbf r_1(t_1)=(2\sin t_1,2\cos t_1,0)$
$C_2:\mathbf r_2(t_2)=(1-t_2)(2,0,0)+t_2(3,3,3)=(2+t_2, 3t_2, 3t_2)$

where $0\le t_1\le\dfrac\pi2$ and $0\le t_2\le1$ . You have

$\mathrm d\mathbf r_1=(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1$
$\mathrm d\mathbf r_2=(1,3,3)\,\mathrm dt_2$

and so the work is given by the integral

$\displaystyle\int_C\mathbf F(x,y,z)\cdot\mathrm d\mathbf r$
$=\displaystyle\int_0^{\pi/2}\mathbf F(2\sin t_1,2\cos t_1,0)\cdot(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1$
${}\displaystyle\,\,\,\,\,\,\,\,+\int_0^1\mathbf F(2+t_2,3t_2,3t_2)\cdot(1,3,3)\,\mathrm dt_2$

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2 years ago

Write a quadratic equation given the roots -1/3 and 5, show your work

Travka [436]

$\boxed{(x - a)(x - b) = 0}$

The equation above is the intercept form. Both a-term and b-term are the roots of equation.

$x = - \frac{1}{3} \\ x = 5$

These are the roots of equation. Therefore we substitute a = - 1/3 and b = 5 in the equation.

$(x + \frac{1}{3} )(x - 5) = 0$

Here we can convert the expression x+1/3 to this.

$x + \frac{1}{3} = 0 \\ 3x + 1 = 0$

Rewrite the equation.

$(3x + 1)(x - 5) = 0$

Simplify by multiplying both expressions.

$3 {x}^{2} - 15x + x - 5 = 0 \\ 3 {x}^{2} - 14x - 5 = 0$

Answer Check

Substitute the given roots in the equation.

$3 {(5)}^{2} - 14(5) - 5 = 0 \\ 75 - 70 - 5 = 0 \\ 75 - 75 = 0 \\ 0 = 0$

$3( - \frac{1}{3} )^{2} - 14( - \frac{1}{3}) - 5 = 0 \\ 3( \frac{1}{9} ) + \frac{14}{3} - 5 = 0 \\ \frac{1}{3} + \frac{14}{3} - \frac{15}{3} = 0 \\ \frac{15}{3} - \frac{15}{3} = 0 \\ 0 = 0$

The equation is true for both roots.

Answer

$\large \boxed {3 {x}^{2} - 14x - 5 = 0}$

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2 years ago