What number is midway between 2/3 and 1?

Answers

Answer 1

Answer: 5/6 is about midway through 2/3 and 1. It helps if you lay everything out in a number line like below:

0______________1/4______________ 2/4______________3/4___________1

0____________________1/3___________________2/3________________ 1

0________1/6________ 2/6________3/4_________4/6________5/6_______1

Additionally, 2/3 = .66 so the distance from 2/3 to 1 is approximately .34. If you divide that by 2 to see what the middle is you get .17, which you then add to .66.

.66 + .17 = .83

.83 = 5/6

Answer 2

Answer:

5/6

Step-by-step explanation:

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Question 12What is the length of a segment with endpoints at (-3, 4) and (4, 4)?

Find the Area of the figure below, composed of a rectangle and a semicircle. Round to the nearest tenths place.

Answers

Answer:

153.1 square unit

Step-by-step explanation:

for rectangle

length = 16

breadth = 8

area of rectangle = l*b

=16*8

=128 square unit

semicircle

diameter =8

area of semicircle =πd^2/8

=3.14 * 8^2 /8

=3.14 *64/8

=200.96/8

=25.12 square unit

area of the givem figure = area of rectangle + area of semicircle

=128 + 25.12

=153.12

=153.1 square unit

Final answer:

To find the area of the figure made up of a rectangle and a semicircle, calculate the areas of both shapes and add them together. Use the formula for the area of a rectangle, and calculate the area of the semicircle by finding half the area of a full circle with the same radius. Finally, sum up the areas to find the total area of the figure.

Explanation:

To find the area of the figure, we need to find the areas of the rectangle and the semicircle, and then add them together. The area of the rectangle is found by multiplying its length by its width. The area of the semicircle is half the area of a full circle with the same radius. Finally, we add the two areas together to get the total area of the figure.

Let's assume the length of the rectangle is 10 units and the width is 5 units. The area of the rectangle is then 10 units * 5 units = 50 units^2. If the radius of the semicircle is 3 units, the area of the semicircle is half the area of a full circle with radius 3 units, which is 1/2 * 3.14 * 3^2 = 14.13 units^2 (rounded to the nearest tenths place).

Therefore, the total area of the figure is 50 units^2 + 14.13 units^2 = 64.13 units^2 (rounded to the nearest tenths place).

Learn more about Area of composite figures here:

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The length and a quarter of the width of a rectangle add up to 7 in, while the length and the width add up to 10 in. What are the rectangle's length and width?

Answers

The length of rectangle is 6 inches and width is 4 inches.

Step-by-step explanation:

Let,

l be the length of rectangle

w be the width of rectangle.

Quarter means $(1)/(4)$ .

According to given statement;

$l+(1)/(4)w=7\nl+0.25w=7\ \ \ Eqn\ 1$

l+w = 10 Eqn 2

Subtracting Eqn 1 from Eqn 2

$(l+w)-(l+0.25w)=10-7\nl+w-l-0.25w=3\n0.75w=3$

Dividing both sides by 0.75

$(0.75w)/(0.075)=(3)/(0.75)\nw=4$

Putting w=4 in Eqn 2

$l+4=10\nl=10-4\nl=6$

The length of rectangle is 6 inches and width is 4 inches.

Keywords: rectangle, linear equation

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Find the diagonal of a square with a perimeter of 28 inches

Answers

Perimeter of a square = 4 x the length of one side = 28 in.
Length of one side = 7 in.
Length of the diagonal = square root of (7² + 7²) = 7√2 = 9.899 in (rounded)

$a-side\ of\ square\n\nperimeter\ is\ 28in\n\n4a=28\ \ \ \ /:4\na=7\ (in)\n\nThe\ diagonal\ (d)\ equal\ a\sqrt2\n\nd=7\sqrt2\ inches.$

What is the surface area of a rectangular prism with a length of 6 in., a width of 4 in., and a height of 10 in.?in2

Answers

6*4*10 = 248 in2 here we go.

The area of a playground is 216 yd^2. the width of the playground is 6 yd longer than its length find the length and width.A. length=18, width=12
B. length=12, width=18
C. length=18, width=24
D.length=24, width=18

Answers

Answer: B. length=12, width=18

Step-by-step explanation:

Let x = Length of the playground then width = x+6

Area of a rectangular playground = Length x width

Since , area of playground = $216\ yd^2.$

Then , we have

$216=(x)*(x+6)\n\n\Rightarrow\ x^2+6x\n\n\Rightarrow\ x^2+6x-216=0\n\n\Rightarrow\ x^2+18x-12x-216=0\n\n\Rightarrow\ x(x+18)-12(x+18)=0\n\n\Rightarrow\ (x-12)(x+18)=0\n\n\Rightarrow\ x=12\text{ or }x=-18$

But length cannot be negative , so reject x= -18 .

Thus , the length of playground = 12 yd

Then , width = 12+6 = 18 yd

Hence, the correct answer is B. length=12, width=18

Let the length of the playground be x, then the width is 6 + x.
Area = length * width = x * (6 + x) = 6x + x^2 = 216

Solving the quadratic equation x^2 + 6x - 216 = 0, we have x = 12 or -18

i.e length = 12 and width = 6 + 18 = 18

X^a+b/x^2a. x^b+c/x^2b. xc+a/x^2c

Answers

Answer:

See the image for the answer... Hope this helps!

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