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3 years ago

Find the common difference of the sequence shown. 1/2, 1/4, 0, ... -1/8 -1/4 -1/2

1 answer:

8 0

Common difference in Arithmetic sequences is defined as the fixed amount referring to the fact that the difference between two successive terms yields the constant value that was added. In order to know the common difference of the given sequence above, we just have to deduct 1/2 from 1/4 and the answer is -1/4. Therefore, the correct answer would be the third option. Hope this answer helps.

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The answer to this question here 0.1x=0.2(x+2) x=-4

Elenna [48]

Answer:

the answer is 76

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

ASAP 30 + Brainliest Please only solve 2 - 5

hichkok12 [17]

QUESTION 2a

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

$Area=6a cm^2$ .

QUESTION 2b

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

$Area=7y$

QUESTION 2c.

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

$Area=8x$

QUESTION 2d

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

$Area=b^2 m^2$

QUESTION 2e

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

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

$Area=12a cm^2$

QUESTION 3a

Perimeter is the distance around the figure.

Let P be the perimeter, then

$P=x+x+x+x$

The expression for the perimeter is

$P=4x mm$

QUESTION 3b

The given figure is a rectangle.

Let P, be the perimeter of the given figure.

$P=L+B+L+B$

This simplifies to

$P=2L+2B$

$P=2(L+B)$

QUESTION 3c

The given figure is a parallelogram.

Perimeter is the distance around the parallelogram

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

This simplifies to,

$Perimeter=6q+2P$

$Perimeter=2(3q+P)$

QUESTION 3d

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,

$P=20b mm$

QUESTION 3e

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,

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

$Perimeter=13m mm$

QUESTION 3g

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,

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

$P=9x+5$

QUESTION 3h

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

6 0

3 years ago

Read 2 more answers

A gardener has 520 feet of fencing to fence in a rectangular garden. One side of the garden is bordered by a river and so it doe

Bond [772]

The shortest side is 130 feet, the longest side is 260 feet and the greatest possible area is 33800 square feet

<h3>What dimensions would guarantee that the garden has the greatest possible area?</h3>

The given parameter is

Perimeter, P = 520 feet

Represent the shorter side with x and the longer side with y

One side of the garden is bordered by a river:

So the perimeter is:

P = 2x + y

Substitute P = 520

2x + y = 520

Make y the subject

y = 520 - 2x

The area is

A = xy

Substitute y = 520 - 2x in A = xy

A = x(520 - 2x)

Expand

A = 520x - 2x^2

Differentiate

A' = 520 - 4x

Set to 0

520 - 4x = 0

Rewrite as:

4x= 520

Divide by 4

x= 130

Substitute x= 130 in y = 520 - 2x

y = 520 - 2 *130

Evaluate

y = 260

The area is then calculated as:

A = xy

This gives

A = 130 * 260

Evaluate

A = 33800

Hence, the shortest side is 130 feet, the longest side is 260 feet and the greatest possible area is 33800 square feet