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3 years ago

Find the base and height

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

6 0

Answer:

not enough information I don't know what g=

Step-by-step explanation:

Harman [31]3 years ago

3 0

Give more information

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Determine the nature of the roots: 5x2 + 20x + 20 - 0

tankabanditka [31]

Two distinct real solutions

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3 years ago

Which associations best describe the data in the table?

IgorC [24]

In order to find out the association, you need to plot the data. Given that m is x, and n is y, after plotting, the graph shows a positive linear association. The graph rises from left to right. So the answer is letter D.

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3 years ago

Triangles T1 and T2 are congruent.

Inessa [10]

The single transformation that can be used to map T₁ onto T₂ is Rotate T₁ 180° about the origin.

<h3>What transformation would map T₁ onto T₂?</h3>

When a shape is rotated 180° about the origin, it means that the points go from (x, y) to (-x, -y).

The points on T₁ if rotated 180° about the origin would go from:
(-2, -2) to (2, 2)

(5, -6) to (-5, 6)

(5,0) to (-5, 0)

These are the points of T₂ so the way to map T₁ on T₂ is to use a 180° about the origin.

Find out more on rotating about the origin at brainly.com/question/16200542

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3 0

2 years ago

Need help ASAP!!

Ilia_Sergeevich [38]

An object that is in motion as a projectile follows a path or trajectory of a parabola

The function and values are;

a) The equation of the quadratic function is; $\underline{y = \dfrac{111}{140} \cdot x - \dfrac{3}{140} \cdot x^2}$

b) The maximum height of the ball is approximately <u>7.334 m</u>

c) Horizontal distance at maximum height <u>18.8 meters</u>

<u />

Reason:

a) Known parameters are;

Let f(x) = a·x² + b·x + c represent the equation of the parabola modelling the path of the ball, we have;

Points on the path of the parabola = (0, 0), (35, 1.5), 37, 0)

Plugging the values gives;

0 = a·0² + b·0 + c

Therefore, c = 0

1.5 = 35²·a + 35·b

0 = 37²·a + 37·b

Solving gives;

a = -3/140, b = 111/140

The equation of the quadratic function is therefore;

$\underline{y = f(x) = \dfrac{111}{140} \cdot x - \dfrac{3}{140} \cdot x^2}$

b) The maximum height is given by the vertex of the parabola

The x-coordinate at the vertex is the point $-\dfrac{b}{2 \cdot a}$

Which gives;

$x-coordinate = \dfrac{\frac{111}{140} }{2 \times \dfrac{3}{140} } = 17.5$

The maximum height is therefore;

$f(x)_{max} = \dfrac{111}{140} \times 17.5 - \dfrac{3}{140} \cdot 17.5^2 \approx 7.334$

The maximum height of the ball is approximately 7.334 m

c) The distance the ball has travelled to horizontally is given by half of the range, <em>R</em> as follows;

The range of the motion, R = 37 meters

$Horizontal \ distance \ to \ maximum \ height = \dfrac{R}{2}$

Therefore;

$Horizontal \ distance \ to \ maximum \ height = \dfrac{37}{2} = 18.5$

The distance the ball has travelled horizontally to reach the maximum height horizontally <u>18.5 meters</u>

Learn more about the trajectory of a projectile here:

brainly.com/question/13646224

8 0

3 years ago

CAN SOMEONE PLEASE HELP ME???

zmey [24]

Step-by-step explanation:

the graph A is represents the ordered pairs.

4 0

3 years ago