Remove parentheses.

-3/8 -1/6

Find the Least Common Denominator (LCD) of 3/8, 1/6. In other words, find the Least Common Multiple (LCM) of 8,6.

LCD=24

Make the denominators the same as the LCD.

- 3x3/8x3 - 1x4/6x4

Simplify. Denominators are now the same

-9/24 - 4/24

Join the denominators

-9-4/24

Simplify

-13/24

7 0

3 years ago

The area of a trapezoid is 216 in2 and it two bases are 14 and 22 inches long. Find the height of the trapezoid.

artcher [175]

Answer:

12 inches

Step-by-step explanation:

The formula for a trapezoid is

A = 1/2(a +b)h

The formula to find the height =

h = 2A/a + b

A= 216 in²

a =14 in

b = 22 in

Hence,

h = 2 × 216/14 + 22

h = 12 inches

The height of the trapezoid = 12 inches

6 0

3 years ago

ANSWER ALL 5 PARTS.

N76 [4]

A function $f$ from a set A to a set B is defined as a relation that assings to each element $x$ in the set A exactly one element $y$ in the set B. The set A is called the domain of the function while the set B is the range. So we have five statements and need to find some functions. Melissa decides to reserve a patch in her vegetable garden for growing bell peppers. If each side of the tomato patch is $x$ feet, then we have a square patch as shown in the Figure below.

1.a) Write the function Wa(x) representing the width of the bell pepper patch.

We know that she wants its width to be half the width of the tomato patch. Let $x$ be the width of the tomato patch, then the function that matches this statement is:

$\boxed{Wa(x)=\frac{x}{2}}$

1.b) Write the function La(x) representing the length of the bell pepper patch.

In this case Melissa wants its length to exceed the length of the tomato patch by 2 feet. To do this we enlarge the length of the tomato patch 2 feet. Therefore the function is the following:

 $\boxed{La(x)=x+2}$

2. Area of the bell pepper patch in terms of x.

Given that the bell pepper patch is a rectangle, then the area of a rectangle is the product of the length and width. So:

 $A=(\frac{x}{2})(x+2) \\ \\ \therefore \boxed{A=\frac{x(x+2)}{2}}$

3. Combined area of the tomato patch and the bell pepper patch.

This function is the sum of both the area of the tomato patch and the bell pepper patch. So:

 $Aar(x)=x^{2}+\frac{x(x+2)}{2} \rightarrow Aar(x)=x^{2}+ \frac{x^{2}}{2}+x \rightarrow Aar(x)=\frac{3x^{2}}{2}+x \\ \\ \therefore \boxed{Aar(x)=\frac{x(3x+2)}{2}}$

4. Write the function Aa(x) for the remaining planting area in the garden.

The remaining planting area in the garden are the rectangles in red. So we need to subtract the width of the bell pepper patch from the width of the tomato patch and multiply it by 2. In mathematical language this is given by:

 $Aa(x)=2(x-\frac{x}{2}) \rightarrow Aa(x)=x$

5. Find the area of the remaining space in the garden after planting tomatoes and bell peppers.

Given that Melissa wants the area of the bell pepper patch to be 31.5 square feet, then it is true that:

 $31.5=\frac{x(x+2)}{2} \rightarrow x^{2}+2x-64=0 \\ \\ solving \ for \ x: \\ x=7.06$

Therefore the area of the remaining space is:

 $\boxed{Aa(7.06)=7.06ft^{2}}$

6 0

4 years ago