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3 years ago

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Classify the function as linear or quadratic and identify the quadratic linear and constant terms. PLEASE HELP ME!!!

1 answer:

patriot [66]3 years ago

5 0

The equation
$y = - 3 {x}^{2} - 29x + 30$
is quadratic, since the highest power of x is x^2. The quadratic term is -3x^2, the linear term is -29x, and the constant is 30.

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ASAP 30 + Brainliest <br><br> Please only solve 2 - 5

hichkok12 [17]

<u>QUESTION 2a</u>

We want to find the area of the given right angle triangle.

We use the formula

$Area=\frac{1}{2}\times base\times height$

The height of the triangle is $=a cm$ .

The base is $12cm$ .

We substitute the given values to obtain,

$Area=\frac{1}{2}\times 12\times a cm^2$ .

This simplifies to get an expression for the area to be

$Area=6a cm^2$ .

<u>QUESTION 2b</u>

The given diagram is a rectangle.

The area of a rectangle is given by the formula

$Area=length \times width$

The length of the rectangle is $l=7cm$ and the width of the rectangle is $w=ycm$ .

We substitute the values to obtain the area to be

$Area=7 \times y$

The expression for the area is

$Area=7y$

<u>QUESTION 2c.</u>

The given diagram is a rectangle.

The area of a rectangle is given by the formula

$Area=length \times width$

The length of the rectangle is $l=2x cm$ and the width of the rectangle is $w=4 cm$ .

We substitute the values to obtain the area to be

$Area=2x \times 4$

The expression for the area is

$Area=8x$

<u>QUESTION 2d</u>

The given diagram is a square.

The area of a square is given by,

$Area=l^2$ .

where $l=b m$ is the length of one side.

The expression for the area is

$Area=b^2 m^2$

<u>QUESTION 2e</u>

The given diagram is an isosceles triangle.

The area of this triangle can be found using the formula,

$Area=\frac{1}{2}\times base\times height$ .

The height of the triangle is $4cm$ .

The base of the triangle is $6a cm$ .

The expression for the area is

$Area=\frac{1}{2}\times 6a \times 4cm^2$

$Area=12a cm^2$

<u>QUESTION 3a</u>

Perimeter is the distance around the figure.

Let P be the perimeter, then

$P=x+x+x+x$

The expression for the perimeter is

$P=4x mm$

<u>QUESTION 3b</u>

The given figure is a rectangle.

Let P, be the perimeter of the given figure.

$P=L+B+L+B$

This simplifies to

$P=2L+2B$

Or

$P=2(L+B)$

<u>QUESTION 3c</u>

The given figure is a parallelogram.

Perimeter is the distance around the parallelogram

$Perimeter=3q+P+3q+P$

This simplifies to,

$Perimeter=6q+2P$

Or

$Perimeter=2(3q+P)$

<u>QUESTION 3d</u>

The given figure is a rhombus.

The perimeter is the distance around the whole figure.

Let P be the perimeter. Then

$P=5b+5b+5b+5b$

This simplifies to,

$P=20b mm$

<u>QUESTION 3e</u>

The given figure is an equilateral triangle.

The perimeter is the distance around this triangle.

Let P be the perimeter, then,

$P=2x+2x+2x$

We simplify to get,

$P=6x mm$

QUESTION 3f

The figure is an isosceles triangle so two sides are equal.

We add all the distance around the triangle to find the perimeter.

This implies that,

$Perimeter=3m+5m+5m$

$Perimeter=13m mm$

<u>QUESTION 3g</u>

The given figure is a scalene triangle.

The perimeter is the distance around the given triangle.

Let P be the perimeter. Then

$P=(3x+1)+(2x-1)+(4x+5)$

This simplifies to give us,

$P=3x+2x+4x+5-1+1$

$P=9x+5$

<u>QUESTION 3h</u>

The given figure is a trapezium.

The perimeter is the distance around the whole trapezium.

Let P be the perimeter.

Then,

$P=m+(n-1)+(2m-3)+(n+3)$

We group like terms to get,

$P=m+2m+n+n-3+3-1$

We simplify to get,

$P=3m+2n-1$ mm

QUESTION 3i

The figure is an isosceles triangle.

We add all the distance around the figure to obtain the perimeter.

Let $P$ be the perimeter.

Then $P=(2a-b)+(a+2b)+(a+2b)$

We regroup the terms to get,

$P=2a+a+a-b+2b+2b$

This will simplify to give us the expression for the perimeter to be

$P=4a+3b$ mm.

QUESTION 4a

The given figure is a square.

The area of a square is given by the formula;

$Area=l^2$

where $l=2m$ is the length of one side of the square.

We substitute this value to obtain;

$Area=(2m)^2$

This simplifies to give the expression of the area to be,

$Area=4m^2$

QUESTION 4b

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=5a cm$ is the length of the rectangle and $w=6cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =5a \times 6$

$Area =30a cm^2$

QUESTION 4c

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=7y cm$ is the length of the rectangle and $w=2x cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =7y \times 2x$

The expression for the area is

$Area =14xy cm^2$

QUESTION 4d

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=3p cm$ is the length of the rectangle and $w=p cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =3p \times p$

The expression for the area is

$Area =3p^2 cm^2$

See attachment for the continuation

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Step-by-step explanation:

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