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3 years ago

A sequence can be generated ny using (look at image below)? someone please help:)

Mathematics

1 answer:

mixer [17]3 years ago

8 0

Answer:

6, 24, 96, 384

Step-by-step explanation:

a₁ = 6

a₂ = 4 x a₁ = 4 x 6 = 24

a₃ = 4 x a₂ = 4 x 24 = 96

a₄ = 4 x a₃ = 4 x 96 = 384

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9. Solve the following system of equations. x + 2y − 6 = z 3y − 2z = 7 4 + 3x = 2y − 5z A. x = 0, y = 5, z = 4 B. C. D. x = 2, y

Inessa [10]

X + 2y - 6 = z..........x + 2y - z = 6
3y - 2z = 7
4 + 3x = 2y - 5z.....3x - 2y + 5z = -4

x + 2y - z = 6.....multiply by -3
3x - 2y + 5z = -4
---------------------
-3x - 6y + 3z = -18 (result of multiplying by -3)
3x - 2y + 5z = -4
--------------------add
-8y + 8z = - 22

-8y + 8z = -22
3y - 2z = 7 .....multiply by 4
----------------
-8y + 8z = -22
12y - 8z = 28 (result of multiplying by 4)
----------------add
4y = 6
y = 6/4 reduces to 3/2

-8y + 8z = -22
-8(3/2) + 8z = -22
-12 + 8z = -22
8z = -22 + 12
8z = -10
z = -10/8 reduces to -5/4

x + 2y - 6 = z
x + 2(3/2) - 6 = -5/4
x + 3 - 6 = -5/4
x - 3 = -5/4
x = -5/4 + 3
x = -5/4 + 12/4
x = 7/4

solution is (7/4, 3/2, -5/4).....x = 7/4, y = 3/2, and z = -5/4

4 0

3 years ago

If {X)= x-9 and d(x) = x + 3, what is the domain of (cd)(x)?

Aleonysh [2.5K]

Answer:

i wish i knew

Step-by-step explanation:

i wanted to type

5 0

3 years ago

find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

6 0

2 years ago

A rectangle has its vertices at (-4, -3), (-4,7), (1,7), (1, -3). What part, in percent, of the rectangle is located in Quadrant

Rasek [7]

Consider the attached figure. The whole rectangle is ABCD, while AEGF is the part located in the third quadrant. In fact, this quadrant is composed by all the points with both coordinates negative.

To answer the question, let's compute the area of the two rectangles and see what part of ABCD is AEGF.

A and B have the same x coordinate, so the length of AB is given by the absolute difference of their y coordinates:

$\overline{AB} = |A_y-B_y| = |-3-7| = |-10| = 10$

Similarly, but exchanging the role of x and y, we compute the length of BC:

$\overline{BC} = |B_x-C_x| = |-4-1| = |-5| = 5$

So, the area of the rectangle is $\overline{AB} \cdot \overline{BC} = 10\cdot 5 = 50$

The same procedure allows us to compute width and height of the sub-rectangle in the third quadrant:

$\overline{AE} = |A_y-E_y| = |-3-0| = |-3| = 3$

$\overline{EG} = |E_x-G_x| = |-4-0| = |-4| = 4$

So, the area of the portion located in the third quadrant is $\overline{AE} \cdot \overline{EG} = 3\cdot 4 = 12$

This means that the ratio between the two area is

$\cfrac{\text{area }AEGF}{\text{area }ABCD} = \cfrac{12}{50}$

If we want this ratio to be a percentage, just make sure that the denominator is 100:

$\cfrac{12}{50} = \cfrac{12}{50}\cdot \cfrac{2}{2} = \cfrac{24}{100} = 24\%$

3 0

2 years ago

What is the greatest common factor of the expression 12x4+18x2

Nata [24]

Answer:

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers