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2 years ago

In 8- 11 find the area of each parrelogram or rhombus

Mathematics

1 answer:

enyata [817]2 years ago

4 0

The areas of the two parallelograms are 105 square inches and 20.997 square yards, respectively.

<h3>How to determine the area of a parallelogram</h3>

A parallelogram is a quadrilateral with two pairs of parallel sides with equal length and two pairs of angles of equal measure. The area of the parallelogram (<em>A</em>), in square yards or square inches, equals the product of its base length (<em>b</em>), in yards, and its height (<em>h</em>), in yards.

The area of each parallelogram is determined afterwards:

<h3>Exercise 10</h3>

<h3>Exercise 11</h3>

The areas of the two parallelograms are 105 square inches and 20.997 square yards, respectively. $\blacksquare$

To learn more on parallelograms, we kindly invite to check this verified question: brainly.com/question/1563728

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Lease help...

m_a_m_a [10]

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4 years ago

Read 2 more answers

Round to the nearest tenth wa hit is the perimeter of the triangle

klio [65]

Answer:

D. 11.8 cm

Step-by-step explanation:

This triangle is a 30°-60°-90° triangle. There is a shortcut to find the lengths of the sides of a 30°-60°-90° triangle. The longest side is the hypotenuse. The longest side is double the shortest side or, the shortest side is half the longest side. Here the longest side is 5, so the short leg (shortest side) is 2.5

The other shortcut is that the longer leg is the

short leg×sqroot3

Here, the short leg is 2.5, so the long leg is 2.5×sqroot3

Perimeter is all the sides added together.

5 + 2.5+ 2.5×sqrt3

Do the times first.

5 + 2.5 + 4.3

= 11.8

6 0

2 years ago

Not sure what to do here can someone please help

kumpel [21]

Answer:

x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

$\sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}$

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true

x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

$u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}$

Solutions to this equation are ...

u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

x = u² -2 = 2² -2

x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.