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2 years ago

What is the area???

Mathematics

1 answer:

Natalka [10]2 years ago

4 0

Answer:

in squared 252

Step-by-step explanation: idk how but

9 times 12= 108

6 times 15=90

6 times 9=54

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PLEASE HELP WILL MARK THE BRAINLIEST

zhuklara [117]

1. Quadratic equations have the highest degree of 2. Therefore, the maximum amount of solutions can be 2 or unreal. The graphs with 2 solutions would cross over the x-axis twice, 1 solution would touch the x-axis once and 0 solutions would not touch the x-axis.

2. Find x and y intercepts of the equation. Calculate points at a constant intervals and plot points on graphing paper. Connect the points.

3. Following the movement of a ball through air? <-- not sure about this one

Hope I helped :)

5 0

2 years ago

Are these triangles congruent? If so, why?

N76 [4]

<h3>Yes the triangles are congruent by ASA axiom or congruency .</h3>

Because one side of triangle is equal and two angles are equal . So option 1 is your answer .

5 0

3 years ago

PLEASE SHOW WORK....

Irina18 [472]

Answer:

Step-by-step explanation:

omg its says you cant have help

8 0

2 years ago

Given: ∆ABC –iso. ∆, m∠BAC = 120°

lidiya [134]

Answer: $P\triangle ADH = (a+b+\sqrt{a^2+b^2} )$ unit

Step-by-step explanation:

Since Here Δ ABC is a isosceles triangle.

Also, $AH\perp BC$ and $HD\perp AC$

Thus, ∠ ADH = 90°

That is, Δ ADH is the right angle triangle.

In which, AD = a unit and HD = b unit

And, By the definition of Pythagoras theorem,

$AH^2 = HD^2 + AD^2$

⇒ $AH^2 = a^2 +b^2$

⇒ $AH = \sqrt{a^2+b^2}$

Since, the perimeter of a triangle = sum of the all sides of the triangle.

Therefore, perimeter of ΔADH = AD + DH + AH

= $a + b + \sqrt{a^2+b^2}$ unit

4 0

3 years ago

If the point (4,-2) is included in a direct variation relationship, which point also belongs in this direct variation? A. (-4,2)

IceJOKER [234]

Answer:

(-4,2) belongs in this direct variation.

Step-by-step explanation:

Let the direct variation relationship is expressed by the equation y = mx ........ (1), where x and y are in direct variation and k is the variation constant.

Now, the point (4,-2) is included in the direct variation relationship, then from equation (1) we get, -2 = 4m

⇒ $m = - \frac{1}{2}$

Therefore, the equation (1) becomes $y = - \frac{1}{2}x$

⇒ x + 2y = 0 ......... (2)

Now, from the given four options only the point (-4,2) satisfies the relation (2).

Hence, (-4,2) belongs in this direct variation. (Answer)

5 0

3 years ago