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1 year ago

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The length of two parallel sides of a trapezium are 5cm and 7cm and the area is 120cm.sq find the distance between the parallel

sides

1 answer:

Alla [95]1 year ago

5 0

The height or the distance between the two parallel sides of the given trapezium is 20 cm.

A trapezium or trapezoid is a two-dimensional geometric shape that has 4 sides, with one pair of parallel sides.

The parallel sides of a trapezium are called bases while the distance between them is the altitude or height.

The formula for the area of a trapezoid is given by

A = (a + b)/2 x h

where a and b are the parallel bases and h is the height.

Using this formula, the height or the distance between the two parallel sides can be computed.

A = (a + b)/2 x h

120 cm^2 = (5 cm + 7 cm) / 2 x h

120 = (12 / 2) x h

120 = 6h

h = 120 / 6

h = 20 cm

For more examples on solving the area of a trapezium, visit brainly.com/question/16904048.

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Pls help me with my math

givi [52]

Answer:

The definition for the given piecewise-defined function is: $\boxed{\displaystyle\sf\ Option\:D:\:\: f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$ .

Step-by-step explanation:

<h3>General Concepts:</h3>

Piecewise-defined functions.
Interval notations.

<h3>What is a piecewise-defined function?</h3>

A piecewise-defined function represents specific rules over different intervals of the domain.

<h3>Symbols used in expressing interval notations:</h3>

Open interval: This means that the endpoint is <em>not</em> included in the interval.

We can use the following symbols to indicate the <u>exclusion</u> of endpoints in the interval:

Left or right parenthesis, "( )" (or both).

Greater than (>) or less than (<) symbols.
Open dot " $\circ$ " is another way of expressing the exclusion of an endpoint in the graph of a piecewise-defined function.

Closed interval: This implies the inclusion of endpoints in the interval.

We can use the following symbols to indicate the <u>inclusion</u> of endpoints in the interval:

Open- or closed brackets (or both), "[ ]."

Greater than or equal to (≥) or less than or equal to (≤) symbols.
Closed circle or dot, "•" is another way of expressing the <em>inclusion</em> of the endpoint in the graph of a piecewise-defined function.

<h2>Determine the appropriate function rule that defines different parts of the domain. </h2>

The best way to determine which piecewise-defined function represents the graph is by observing the <u>endpoints</u> and <u>orientation</u> of both partial lines.

Open circle on (-1, 2): The graph shows that one of the partial lines has an <em>excluded</em> endpoint of (-1, 2) extending towards the <u>right</u>. This implies that its domain values are defined when x > -1.

Closed circle on (-1, 1): The graph shows that one of the partial lines has an <em>included</em> endpoint of (-1, 1) extended towards the <u>left</u>. Hence, its domain values are defined when x ≤ -1.

Based on our observations from the previous step, we can infer that x > -1 or x ≤ -1 apply to piecewise-defined functions A or D. However, only one of those two options represent the graph.

<h2>Solution:</h2><h3>a) Test option A:</h3>

$\boxed{\displaystyle\sf Option\:A)\:\:\:f(x) = \begin{cases}\displaystyle\sf\ 2x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$

<h3>Piece 1: If x ≤ -1, then it is defined by f(x) = 2x + 2. </h3>

We must choose a domain value that falls within the interval of x ≤ -1 whose output is included is included in the graph of the partial line with a <u>closed dot</u>.

Substitute x = -2 into f(x) = 2x + 2:

f(x) = 2x + 2

f(-2) = 2(-2) + 2
f(-2) = -4 + 2
f(-2) = -2 ⇒ <em>False statement</em>.

⇒ The output value of f(-2) = -2 is <u>not</u> included in the graph of the partial line whose endpoint is at (-1, 1).

<h3>Piece 2: If x > -1, then it is defined by f(x) = x + 4. </h3>

We must choose a domain value that falls within the interval of x > -1 whose output is included in the graph of the partial line with an <u>open dot</u>.

Substitute x = 0 into f(x) = x + 4:

f(x) = x + 4
f(0) = (0) + 4
f(0) = 4 ⇒ <em>True statement</em>.

⇒ The output value of f(0) = 4 <u>is</u> included in the graph of the partial line whose endpoint is at (-1, 2).

Conclusion for Option A:

Option A is not the correct piecewise-defined function because one of the pieces, f(x) = 2x + 2, does not specify the interval (-∞, -1].

<h3>b) Test option D:</h3>

$\boxed{\displaystyle\sf Option\:D)\:\:\:f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$

<h3>Piece 1: If x ≤ -1, then it is defined by f(x) = x + 2. </h3>

We must choose a domain value that falls within the interval of x ≤ -1 whose output is included is included in the graph of the partial line with a <u>closed dot</u>.

Substitute x = -2 into f(x) = x + 2:

f(x) = x + 2
f(-2) = (-2) + 2
f(-2) = 0 ⇒ <em>True statement</em>.

⇒ The output value of f(-2) = 0 <u>is</u> included the graph of the partial line whose endpoint is at (-1, 1).

<h3>Piece 2: If x > -1, then it is defined by f(x) = 2x + 4.</h3>

We must choose a domain value that falls within the interval of x > -1 whose output is included is included in the graph of the partial line with an <u>open dot</u>.

Substitute x = 0 into f(x) = 2x + 4:

f(x) = 2x + 4
f(0) = 2(0) + 4
f(0) = 0 + 4 = 0 ⇒ <em>True statement</em>.

⇒ The output value of f(0) = 4 <u>is</u> included in the graph of the partial line whose endpoint is at (-1, 2).

<h2>Final Answer: </h2>

We can infer that the piecewise-defined function that represents the graph is:

$\boxed{\displaystyle\sf\ Option\:D:\:\: f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$ .

________________________________________

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8 0

2 years ago

PLEASE IVE POSTED THIS FOUR TIMES CAN SOMEONE PLEASE HELP ME WRITE AN EQUATION FOR THIS! I GIVE BRAINLIST!

borishaifa [10]

Answer:

y= -2/3x-3

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What value must k take in order for the following expression to be greater than zero?

svetlana [45]

Basically anything greater than zero...k>0

8 0

2 years ago

Subtract and simplify. Show work please!

ra1l [238]

1) Final expression: $+7y^2-6y-17$

2) Final expression: $21x^3-46x^2+59x-30$

Step-by-step explanation:

1)

The first expression is

$(-2y^2-11y-7)-(-9y^2-5y+10)$

First, we remove the 2nd bracket by changing the sign of all the terms inside:

$-2y^2-11y-7-+9y^2+5y-10$

Now we group the terms with same degree together:

$(-2y^2+9y^2)+(-11y+5y)+(-7-10)$

Now we solve the expression in each brackets:

$(-2y^2+9y^2)+(-11y+5y)+(-7-10)=\\+7y^2-6y-17$

So, this is the final expression.

2)

The second expression is

$(7x-6)(3x^2-4x+5)$

We apply the distributive property, so we rewrite the expression as follows:

$(7x-6)(3x^2-4x+5)=7x(3x^2-4x+5)-6(3x^2-4x+5)$

Solving both brackets,

$7x(3x^2-4x+5)-6(3x^2-4x+5)=\\21x^3-28x^2+35x-18x^2+24x-30$

Now we group terms of same degree together:

$21x^3-28x^2+35x-18x^2+24x-30=\\21x^3+(-28x^2-18x^2)+(35x+24x)-30$

And solving each bracket,

$21x^3+(-28x^2-18x^2)+(35x+24x)-30=\\21x^3-46x^2+59x-30$

So, this is the final expression.

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3 0

3 years ago

The general form of a circle is given as x^2+y^2+4x - 12y + 4 = 0. A) What are the coordinates of the center of the circle? B) W

Alex_Xolod [135]

Answer:

center = (-2, 6)

radius = 6

Step-by-step explanation:

<u>Equation of a circle</u>

$(x-a)^2+(y-b)^2=r^2$

(where (a, b) is the center and r is the radius)

Therefore, we need to rewrite the given equation into the standard form of an equation of a circle.

Given equation:

$x^2+y^2+4x-12y+4=0$

Collect like terms and subtract 4 from both sides:

$\implies x^2+4x+y^2-12y=-4$

Complete the square for both variables by adding the square of half of the coefficient of $x$ and $y$ to both sides:

$\implies x^2+4x+\left(\dfrac{4}{2}\right)^2+y^2-12y+\left(\dfrac{-12}{2}\right)^2=-4+\left(\dfrac{4}{2}\right)^2+\left(\dfrac{-12}{2}\right)^2$

$\implies x^2+4x+4+y^2-12y+36=-4+4+36$

Factor both variables:

$\implies (x+2)^2+(y-6)^2=36$

Therefore:

center = (-2, 6)
radius = √36 = 6

7 0

1 year ago