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3 years ago

Last one on test Parallelogram ABCD is a rectangle.

Mathematics

2 answers:

andreyandreev [35.5K]3 years ago

7 0

Answer: 10

Step-by-step explanation:

2AX=BD

2(3y-5)=5y

6y-10=5y

add 10 to each side 6y+10-10=5y+10

6y=5y+10

subtract 5y from each side 6y-5y=5y-5y+10

6y-5y=y

y=10

marusya05 [52]3 years ago

3 0

Answer: 10

2AX=BD
2(3y-5)=5y
6y-10=5y
add 10 to each side 6y+10-10=5y+10
6y=5y+10
subtract 5y from each side 6y-5y=5y-5y+10
6y-5y=y
y=10

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WHAT IS X³-27 SIMPLIFIED

Eduardwww [97]

Answer:

It is (x - 3)³ - 9x(3 - x)

Step-by-step explanation:

Express 27 in terms of cubes, 27 = 3³:

$= {x}^{3} - {3}^{3}$

From trinomial expansion:

${(x - y)}^{3} = (x - y)(x - y)(x - y) \\$

open first two brackets to get a quadratic equation:

${(x - y)}^{3} = ( {x}^{2} - 2xy + {y}^{2} )(x - y)$

expand further:

${(x - y)}^{3} = {x}^{3} - y {x}^{2} - 2y {x}^{2} + 2x {y}^{2} + x {y}^{2} - {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - {y}^{3} + 3x {y}^{2} - 3y {x}^{2} \\ {(x - y)}^{3} = {x}^{3} - {y}^{3} + 3xy(y - x) \\ \\ { \boxed{( {x}^{3} - {y}^{3} ) = {(x - y)}^{3} - 3xy(y - x)}}$

take y to be 3, then substitute:

$( {x}^{3} - 3^3) = {(x - 3)}^{3} - 9x(3 - x)$

5 0

2 years ago

Please answer the question to the best of your ability :)

Korolek [52]

Answer:

7x^2y(2x-3y)

Step-by-step explanation:

Factor 7x^2y out of 14x^3y-21x^2y^2

3 0

3 years ago

Geometry help. Inscribing circle help and constructing lines.

prohojiy [21]

<h3>1. Inscribe a circle ΔJKL. Identify the point of concurrency that is the center of the circle you drew. </h3>

To inscribe a circle in the attached triangle, we must follow the following steps. The resultant figure is the first one below.

1. Draw the angle bisector of angle JKL. The straight line in the green one.

2. Draw the angle bisector of angle KLJ. The straight line in the blue one.

3. Draw point D at these two lines.

4. Draw a line through point D perpendicular to side JK. The straight line in the red one.

5. Mark point G at the red line and side JK.

6. Draw a circle with center at D (point of concurrency) and a radius of DG.

<h3>2. Circumscribe a circle ΔFGH. Identify the point of concurrency that is the center of the circle you drew. </h3>

To circumscribe a circle in the attached triangle, we must follow the following steps:

1. Draw the perpendicular bisector of the side FG. The straight line in the red one. The resultant figure is the second one below.

2. Draw the perpendicular bisector of the side GH. The straight line in the blue one.

3. The center of the circle is the intersection of these two lines

4. Draw a circle with center at D (point of concurrency).

<h3>3. Construct the two lines tangents to circle</h3>

A tangent to a circle is a straight line that touches the circle at an only point. If a point A is outside a circle, we can draw two lines that pass through this point and are tangent to the circle at two different points each. So this is shown in the third figure. Lines red and blue are tangent to the circle at two different points.

3 0

2 years ago

If n is a positive integer, then lim n→∞ 1/n[(1/n)^2+(2/n)^2+...+((n-1)/n)^2] can be expressed as...

Svetach [21]

Answer:

you ended up having = 1/n^3[1^2+2^2+3^2...(n-1)^2]

hope this helps :')

7 0

2 years ago

I have corner points of:

WARRIOR [948]

The points you found are the vertices of the feasible region. I agree with the first three points you got. However, the last point should be (25/11, 35/11). This point is at the of the intersection of the two lines 8x-y = 15 and 3x+y = 10

So the four vertex points are:
(1,9)
(1,7)
(3,9)
(25/11, 35/11)

Plug each of those points, one at a time, into the objective function z = 7x+2y. The goal is to find the largest value of z

------------------

Plug in (x,y) = (1,9)
z = 7x+2y
z = 7(1)+2(9)
z = 7+18
z = 25
We'll use this value later.
So let's call it A. Let A = 25

Plug in (x,y) = (1,7)
z = 7x+2y
z = 7(1)+2(7)
z = 7+14
z = 21
Call this value B = 21 so we can refer to it later

Plug in (x,y) = (3,9)
z = 7x+2y
z = 7(3)+2(9)
z = 21+18
z = 39
Let C = 39 so we can use it later

Finally, plug in (x,y) = (25/11, 35/11)
z = 7x+2y
z = 7(25/11)+2(35/11)
z = 175/11 + 70/11
z = 245/11
z = 22.2727 which is approximate
Let D = 22.2727

------------------

In summary, we found
A = 25
B = 21
C = 39
D = 22.2727

The value C = 39 is the largest of the four results. This value corresponded to (x,y) = (3,9)

Therefore the max value of z is z = 39 and it happens when (x,y) = (3,9)

------------------

Final Answer: 39

7 0

3 years ago