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2 years ago

5

A square has a side of length of 12 centimeters. Which of he following is closest to the length of a diagonal of the square?

1 answer:

dimaraw [331]2 years ago

4 0

<span>length of a diagonal of the square = </span>√(12^2 + 12^2)
= √(144 + 144)
= √288
= 16.97 cm

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Find the midpoint of AB for: A(7,0)|and B(0,3)

liberstina [14]

Answer:

$\boxed {\boxed {\sf (\frac {7}{2}, \frac{3}{2}) \ or \ (3,5, 1.5) }}$

Step-by-step explanation:

The midpoint is essentially a point with the average of the 2 x-coordinates and the 2 y-coordinates.

The formula is:

$(\frac {x_1+x_2}{2}, \frac{y_1+y_2}{2})$

We are given two points: A (7,0) and B (0, 3). Remember points are written as (x, y).

Therefore,

$x_1= 7 \\y_1=0 \\x_2=0 \\x_2=3$

Substitute the values into the formula.

$(\frac {7+0}{2}, \frac{0+3}{2})$

Solve the numerators first.

$(\frac {7}{2}, \frac{3}{2})$

The midpoint can be left like this because the fractions are reduced, but it can be written as decimals too.

$(3.5, 1.5)$

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3 years ago

When this open-ended cylinder is opened out, it forms a rectangle with a width of 25 cm. What is the area of the rectangle?.

nalin [4]

Answer:

Just want the points

Step-by-step explanation:

4 0

2 years ago

Given the system of equations, match the following items. x + 2y = 7 x - 2y = -1 y determinant 2. x determinant 3. system determ

Veseljchak [2.6K]

Answer:

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3 years ago

Read 2 more answers

A rancher wants to fence in an area of 1500000 square feet in a rectangular field and then divide it in half with a fence down t

VLD [36.1K]

Answer:

6000 ft

Step-by-step explanation:

Let length of rectangular field=x

Breadth of rectangular field=y

Area of rectangular field= $1500000$ square ft

Area of rectangular field= $l\times b$

Area of rectangular field= $x\time y$

$1500000=xy$

$y=\frac{1500000}{x}$

Fencing used ,P(x)= $x+x+y+y+y=2x+3y$

Substitute the value of y

P(x)= $2x+3(\frac{1500000}{x})$

$P(x)=2x+\frac{4500000}{x}$

Differentiate w.r.t x

$P'(x)=2-\frac{4500000}{x^2}$

Using formula: $\frac{dx^n}{dx}=nx^{n-1}$

$P'(x)=0$

$2-\frac{4500000}{x^2}=0$

$\frac{4500000}{x^2}=2$

$x^2=\frac{4500000}{2}=2250000$

$x=\sqrt{2250000}=1500$

It is always positive because length is always positive.

Again differentiate w.r.t x

$P''(x)=\frac{9000000}{x^3}$

Substitute x=1500

$P''(1500)=\frac{9000000}{(1500)^3}>0$

Hence, fencing is minimum at x=1 500

Substitute x=1 500

$y=\frac{1500000}{1500}=1000$

Length of rectangular field=1500 ft

Breadth of rectangular field=1000 ft

Substitute the values

Shortest length of fence used= $2(1500)+3(1000)=6000 ft$

Hence, the shortest length of fence that the rancher can used=6000 ft

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3 years ago

*NEED HELP A.S.A.P*

Greeley [361]

The answer would be A

8 0

3 years ago