12. What is the area of the quadrilateral?

Mathematics

1 answer:

ivolga24 [154]2 years ago

4 0

The answer is A 100 square centimeters

You might be interested in

Use the equation below to find v, if u= 12, a=9, and t=4.<br> v=u+at

Lina20 [59]

Answer:

v=48

Step-by-step explanation:

$v=?\\\\u=12\\\\a=9\\\\t=4\\\\v=u+at\\\\v=(12)+(9)(4)\\\\v=12+36\\\\v=48$

5 0

3 years ago

During a recent eruption, the volcano spewed out copious amount of ash. One small piece of ash was ejected from the volcano with

nlexa [21]

For question number 1:The plot H = H(t) is the parabola and it reaches its maximum in the moment when exactly at midpoint between the roots t = 0 and t = 23. At that moment t = 23/2 or 11.5 seconds.
For question number 2:To find the maximal height, just simply substitute t = 11.5 into the quadratic equation. The answer would be 22.9.
For question number 3:H(t) = 0, or, which is the same as -16t^2 + 368t = 0.Factor the left side to get -16*t*(t - 23) = 0.t = 0, relates to the very start of the process, when the ash started its way up.The other root is t = 23 seconds, and it is precisely the time moment when the bit of ash will go back to the ground.

6 0

3 years ago

Determine whether each ordered pair is a solution to the inequality x+y<−1.

leva [86]

Answer:

Step-by-step explanation:

coordinates pair (2,11) does not go with the inequality

6 0

3 years ago

Plz help me well mark brainliest if correct....??

oee [108]

Answer:

Table B

Step-by-step explanation:

6 0

3 years ago

What is the center of the circle that you can circumscribe about a triangle with vertices A(2, 6), B(2, 0), and C(10, 0)?

denis23 [38]

The three points A,B,C are all points on this circle.
Each point is then equal distance from the center, that distance being the radius of the circle.
Using the distance formula, we can find the center of the circle (x,y):
$d^2 = (x-x_0)^2 + (y-y_0)^2$
Plugging in points A and B into distance formula, then setting them equal to each other gives:
$(x-2)^2+(y-6)^2 = (x-2)^2 + y^2$
Right away we can cancel out the x terms leaving:
$(y-6)^2 = y^2$
Expand Left side and Solve for y:
$y^2 -12y +36 = y^2$
$y = 3$
Plug in points B and C as before:
$(x-2)^2 + y^2 = (x-10)^2 + y^2$
Here we can cancel the y-terms.
Expand and solve for x:
$x^2 -4x+4 = x^2 -20x+100$
$16x = 96$
$x = 6$
Therefore the center of the circle is the point (6,3)

6 0

3 years ago