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2 years ago

One side of a square garden is 8 feet long. How can you find the area of That is the the garden?

Mathematics

1 answer:

Annette [7]2 years ago

6 0

Answer:

64 ft

Step-by-step explanation:

A square has to be equal on all 4 sides for it to be considered a square. Considering that it is 8 feet on one side it would be 8 feet on all the others as well. From there you multiply 8*8 and receive the solution of 84 feet for the area of the garden.

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You want to place a towel bar that is 10 1⁄4 inches long in the center of a door that is 26 1⁄3 inches wide. How far should you

attashe74 [19]

The bar should be 8 1/24 from the each edge of the door.

We need to subtract 10 1/4 from 26 1/3 to get the fraction of the space not covered by the towel bar.
We also need to divide the difference by 2 because we placed the towel bar in the center of the door.

1st we need to convert the mixed fractions into fractions to perform subtraction.
26 1/3 = ((26*3)+1)/3 = 79/3
10 1/4 = ((10*4)+1)/4 = 41/4

Steps in Subtracting Fractions
Step 1. Make sure the denominator is the same. 3 and 4 are the denominators, they are not the same but they are factor of 12. So,
79/3 must be multiplied by 4 = 79 * 4 / 3 * 4 = 316 / 12
41/4 must be multiplied by 3 = 41 * 3 / 4 * 3 = 123 / 12

Step 2. Subtract the numerators and place them above the common denominator
316/12 - 123/12 = 316 - 123 / 12 = 193 / 12

Before we can simplify the fraction, we must divide it by two to get the measurement of each edge of the door.

Steps in dividing fractions.
Step 1. Get the reciprocal of the 2nd fraction.
1st fraction : 193 / 12
2nd fraction : 2 /1 ⇒ reciprocal 1/2

Step 2. Multiply the 1st fraction to the reciprocal of the 2nd fraction
193 / 12 * 1/2 = 193 * 1 / 12 * 2 = 193 / 24

Step 3. Simplify the fraction.
193 / 24 = 8 1/24

8 0

3 years ago

How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%

blondinia [14]

The trick is to exploit the difference of squares formula,

$a^2-b^2=(a-b)(a+b)$

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

$(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2$

Whatever you do to the denominator, you have to do to the numerator too. So

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2$

Expand the numerator:

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}$

So we have

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2$

But √12 = √(3•4) = 2√3, so

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}$

7 0

3 years ago

In triangular shape how many angles are greater than a right angle

boyakko [2]

A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted. In Euclidean geometry any three points, when non-collinear, determine a unique triangle and a unique plane. This article is about triangles in Euclidean geometry except where otherwise noted

5 0

3 years ago

Read 2 more answers

I need some help with math

Irina18 [472]

Answer:

$(y-7)^2+(x-2)^2=16$

and

$(x+2)^2+(y-15)^2 = 9$

Step-by-step explanation:

The standard equation of a circle is $(x-h)^2+(y-k)^2=r^2$ where the coordinate (h,k) is the center of the circle.

Second Problem:

We can start with the second problem which uses this info very easily.
(h,k) in this problem is (-2,15) simply plug these into the equation. $(x--2)^2+(y-15)^2=r^2$ .
We can also add the radius 3 and square it so it becomes 9. The equation.
This simplifies to $(x+2)^2+(y-15)^2 = 9$ .

First Problem:

The first problem takes a different approach it is not in standard form. But we can convert it to standard form by completing the square.
$y^2-14y+x^2-4x+37=0$ first subtract 37 from both sides so the equation is now $y^2-14y+x^2-4x=-37$ .
$y^2-14y+x^2-4x+37=0$ by adding $(-\frac{b}{2a} )^2$ to both the x and y portions of this equation you can complete the squares. $(-\frac{b}{2a})^2=(-\frac{-14}{2(1)})^2$ and $(-\frac{-4}{2(1)})^2$ which equals 49 and 4.
Add 49 and 4 to both sides and the equation is now: $y^2-14y+49+x^2-4x+4=-37+49+4$ You can simplify the y and x portions of the equations into the perfect squares or factored form $(y-7)^2$ and $(x-2)^2$ .
Finally put the whole thing together. $(y-7)^2+(x-2)^2=16$ .