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katrin2010 [14]

3 years ago

7

If I have a circle that is 12 in in diameter and it's cold and a half and then folded in half again what is the area of the resu

lting shape and what is how do I write out the equation

1 answer:

nika2105 [10]3 years ago

3 0

Answer:

<h2>A= πr^2/4</h2>

Step-by-step explanation:

Step one:

given data

diameter of circle= 12 in

the radius is r=d/2=6in

the area of the circle is

A=πr^2

A=3.142*6^2

A=3.142*36

A=113.1in^2

If the circle is folded in half, the area will be A/2= 113.1/2= 56.55in^2

In the corresponding shape is folded again the area is A/4= 28.3in^2

<u>The expression for the last shape is </u>

<u>A= πr^2/4</u>

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(1/2)^(x-3) = (1/4)^(x-5)<br> Solve for x using logs

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(X-3)*log(1/2)=(x-5)*log(1/4)
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3 years ago

Explain how you would graph the line containing a slope of –1/5 that goes through the point (1,–4).

Anettt [7]

Slope = -1/5
Coordinate = (1, -4)

We know, y - y1 = m(x - x1)
y + 4 = -1/5 (x - 1)
y + 4 = -1/5x + 1/5
y = -1/5x -19/5

When, x = 1, y = -1/5(1) - 19/5 = -20/5 = -4
x = 2, y = -1/5(2) - 19/5 = -21/5
x = 3, y = -1/5(3) - 19/5 = -22/5

Here, Your Coordinates would be: (0, -4), (1, -21/5), (2, -22/5)
Mark them & draw the lines. Graph is done!

Hope this helps!

5 0

3 years ago

Read 2 more answers

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

6 0

3 years ago

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Find the value of c so that (x-5) is a factor of the polynomial p(x)

liubo4ka [24]

I think the question is

Find the value of c so that (x-5) is a factor of the polynomial

$p(x) = x^3 + 2x^2 + cx + 10$

The other factor is going to be some quadratic. We can say a few things about its coefficients but let's start by saying in general it's

$q(x)= ax^2 + bx + k$

$p(x) = (x-5)q(x)$

$x^3 + 2x^2 + cx + 10 = (x-5)(ax^2 + bx+k) = ax^3 + (b-5a)x^2 + (k-5b)x - 5k$

Equating respective coefficients,

$a=1$

$b-5a = 2$

$k - 5b = c$

$-5k = 10$

so we get

$b = 2 + 5 = 7$

$k = 10/-5 = -2$

$c = k - 5b = 2 - 5(7)= -37$

Answer: -37

Check:

$(x^2 + 7x - 2)(x - 5) = x^3 + 2 x^2 - 37 x + 10\quad\checkmark$

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Tina and maryann each play tennis. Maryann has won 5 more games than Tina. Is it possible for tina to have won games if the sum

Soloha48 [4]

Answer:

Tina won 5 games more than maryann, and the sum of their games is 29.. 29-5/2= 12

Maryann won 12 games while Tina won 12+5= 17 games.

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3 years ago

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