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Lemur [1.5K]
2 years ago
10

In the given figure ABCD is a trapezoid quadrilateral in quadrilateral AB || DC if OA = 2x+4 cm , OB = 9 x -21 cm , OC = 2x-1 cm

and OD= 3 if so x = ?
find value of x

please answer fast it is urgent class 12 th subject maths ​

Mathematics
2 answers:
Dovator [93]2 years ago
6 0

Answer:

← SOLUTION →

OA OC

___ = _____

OB OD

→ 2x + 4

_______

9x - 21

= 2x -1

<h3>______. ( GIVEN ) </h3>

3

= 3(2x+4)

= (9x - 21) ( 2x -1)

= 6x+12

= 18x² -51 x +21

<h3>= 0</h3>

= 6x² - 19x +3.

= 0

= 6x² -18 x- x +3

<h3>= 0 </h3>

→ 6x ( x-3) - (x - 3)

<h3>= 0 </h3>

→ ( x - 3 ) (6x - 1 )

<h3>= 0 </h3>

<h3>x = 3. or x = 1 </h3>

__

6

[ but 2x-1 = {2 × 1

__ - 1 } < 0

. 6

= x= 3

or

x= 1

___

6

<u>SO </u><u>THE </u><u>VALUE</u><u> OF</u><u> </u><u>X </u><u> </u><u>IS </u><u>=</u><u> </u><u>3</u><u> </u><u>or </u><u>1</u><u>/</u><u>6</u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u>X </u><u>=</u><u> </u><u>3</u>

<u>hope</u><u> it's</u><u> helpful</u><u> to</u><u> </u><u>you </u><u>good</u><u> </u><u>luck</u><u> </u><u>for </u><u>study</u>

kupik [55]2 years ago
6 0

Answer:

<u>x = 1/6 or x = 3</u>

Step-by-step explanation:

Given :

  1. AB ║ DC
  2. OA = (2x + 4) cm
  3. OB = (9x - 21) cm
  4. OC = (2x - 1) cm
  5. OD = 3

Solving :

  • As the sides are parallel, the diagonals are proportional to each other
  • OA/OC = OB/OD
  • (2x + 4)/(2x - 1) = (9x - 21)/3
  • (2x + 4)/(2x - 1) = 3(3x - 7)/3
  • (2x + 4) = (3x - 7)(2x - 1)
  • 2x + 4 = 6x² - 14x - 3x + 7
  • 6x² - 17x - 2x + 3 = 0
  • 6x² - 19x + 3 = 0
  • 6x² - 18x - x + 3 = 0
  • 6x(x - 3) - 1(x - 3) = 0
  • (6x - 1)(x - 3) = 0
  • <u>x = 1/6 or x = 3</u>
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I really need help please answer
belka [17]

Answer: Choice B) between 6:48 AM and 5:12 PM

=========================================================

Explanation:

I've seen this problem before, and the first part states that if the temperature is above 25 degrees C, then the air conditioner turns on to start cooling the castle.

We want to find values of t that satisfy f(t) > 25

This is the same as trying to solve f(t) - 25 > 0

So we have,

f(t) = 24 - 5\cos\left(\frac{\pi}{12}t\right)\\\\   f(t)-25 = 24 - 5\cos\left(\frac{\pi}{12}t\right)-25\\\\   f(t)-25 = -1 - 5\cos\left(\frac{\pi}{12}t\right)\\\\   f(t)-25 > 0\\\\ -1 - 5\cos\left(\frac{\pi}{12}t\right) > 0\\\\

To solve this inequality, we can graph using Desmos as I've done so below. See the attached image.

Note how I used x in place of t. This is because the graphing calculator graphs (x,y) coordinates.

Furthermore, note how I marked the two points (6.769, 0) and (17.231, 0)

Between these two points is when the curve is above the x axis, when f(t) > 25 is true. This is when the temperature is above 25 degrees C, and when the air conditioner is working.

When t = 6.769, this means that roughly 6.769 hours have passed since midnight which means we're somewhere between 6 am and 7 am. The only value that works from the answer choices is 6:48 AM.

Note how 0.769*60 = 46.14 is fairly close to 48. This error is likely due to how things are rounded.

Then focusing on the point  (17.231, 0) means that t = 17.231. So 17.231 hours have elapsed since midnight and we're now at roughly 1700 hours

1700 hours in military time = 17-12 = 5 pm

The time value t = 17.231 is between 5 pm and 6 pm. The only thing that works is 5:12 pm.

We can then note how 0.231*60 = 13.86 which is also probably due to rounding error (it's fairly close to 12 though).

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

In short, we found the function for f(t)-25 and graphed it as shown below. The two points (6.769, 0) and (17.231, 0) help us see when the AC is turned on, which is roughly between 6:48 AM and 5:12 PM. This seems like a reasonable time for the AC to be operational. Something like between 5:12 PM and 6:48 AM seems like a bad idea because the AC would be working mostly at night, instead of during the day.

Anything beyond t = 24 represents the next day, which is when the cycle starts all over again. So we only need to focus on the window from t = 0 to t = 24. Specifically, the portion of the f(t)-25 curve that is above the x axis.

8 0
3 years ago
What is the union of the set (1, 4, 6, 8 1/2,10) and (1, 3, 11/2, 6, 10 1/2)
makkiz [27]

Answer:

(1, 4, 6, 8, \frac{1}{2}, 10, 3, \frac{11}{2})

Step-by-step explanation:

We need to find union of given sets (1, 4, 6, \frac{1}{2}, 10) and (1, 3, \frac{11}{2}, 10, \frac{1}{2})

We know by definition that union of two given sets say A and B is set C with all the unique and common values given in set A and B.

∴ (1, 4, 6, \frac{1}{2}, 10) ∪ (1, 3, \frac{11}{2}, 10, \frac{1}{2}) = (1, 4, 6, 8, \frac{1}{2}, 10, 3, \frac{11}{2})

4 0
3 years ago
3x-6y=18 intercept form
Veronika [31]
3x-6y=18

We need to get y by itself so first, subtract 3x from both sides

-6y= -3x+18

Now divide both sides by -6

y= 1/2x-3
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Evaluate the following expression In e^e.<br> A) e^2<br> B) 1 <br> C) 0<br> D) e
saw5 [17]

Answer:

D) e

Step-by-step explanation:

ln e^e

We know that ln x^a is the same as a ln x

e ln (e)

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e

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Read 2 more answers
The equation r(t) = sin(4t)i + cos(4t)j​, 0t≥0 describes the motion of a particle moving along the unit circle. Answer the follo
lorasvet [3.4K]

Answer:

a) Particle has a constant speed of 4, b) Velocity and acceleration vector are orthogonal to each other, c) Clockwise, d) False, the particle begin at the point (0,1).

Step-by-step explanation:

a) Let is find first the velocity vector by differentiation:

\vec v = \frac{dr_{x}}{dt} i + \frac {dr_{y}}{dt} j

\vec v = 4\cdot \cos 4t\, i - 4 \cdot \sin 4t \,j

\vec v = 4 \cdot (\cos 4t \, i - \sin 4t\,j)

Where the resultant vector is the product of a unit vector and magnitude of the velocity vector (speed). Velocity vector has a constant speed only if magnitude of unit vector is constant in time. That is:

\|\vec u \| = 1

Then,

\| \vec u \| = \sqrt{\cos^{2} 4t + \sin^{2}4t  }

\| \vec u \| = \sqrt{1}

\|\vec u \| = 1

Hence, the particle has a constant speed of 4.

b) The acceleration vector is obtained by deriving the velocity vector.

\vec a = \frac{dv_{x}}{dt} i + \frac {dv_{y}}{dt} j

\vec a = 16\cdot (-\sin 4t \,i -\cos 4t \,j)

Velocity and acceleration are orthogonal to each other only if \vec v \bullet \vec a = 0. Then,

\vec v \bullet \vec a = 64 \cdot (\cos 4t)\cdot (-\sin 4t) + 64 \cdot (-\sin 4t) \cdot (-\cos 4t)

\vec v \bullet \vec a = -64\cdot \sin 4t\cdot \cos 4t + 64 \cdot \sin 4t \cdot \cos 4t

\vec v \bullet \vec a = 0

Which demonstrates the orthogonality between velocity and acceleration vectors.

c) The particle is rotating clockwise as right-hand rule is applied to model vectors in 2 and 3 dimensions, which are associated with positive angles for position vector. That is: t \geq 0

And cosine decrease and sine increase inasmuch as t becomes bigger.

d) Let evaluate the vector in t = 0.

r(0) = \sin (4\cdot 0) \,i + \cos (4\cdot 0)\,j

r(0) = 0\,i + 1 \,j

False, the particle begin at the point (0,1).

7 0
2 years ago
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