If one serving is 1/6 of a tray of lasagna, how many servings are in three trays of lasagna?

Ming used the calculations shown to find out how much he would spend on 6 pounds of ground beef if 10 pounds of ground beef cost

Angelina_Jolie [31]

Answer:

He determined pounds per dollar by dividing 10 by 25 but wrote the unit rate as a dollar value.

Step-by-step explanation:

Given

$\frac{10\ pounds}{\$25} = \frac{10\ pounds}{25} / \frac{\$25}{25} = \frac{0.4}{1}$

$Unit\ Price = 40\ cents$

$40\ cents * 6 = \$2.40$

Required

Determine Ming's error

Ming's error is from here

$\frac{10\ pounds}{\$25} = \frac{10\ pounds}{25} / \frac{\$25}{25} = \frac{0.4}{1}$

He calculated the unit rate as pound per dollar.

So, after calculating the unit rate, the unit should be:

$Unit\ Rate = 0.4\ pound/\$$

But instead, he solved as:

$Unit\ Rate = \$0.4/ pound$

Hence, (a) is correct

5 0

2 years ago

Read 2 more answers

The point-slope form of the equation of the line that passes through (–4, –3) and (12, 1) is y – 1 = y minus 1 equals StartFract

Zina [86]

Answer:

The equation of the line in standard form is

$x-4y=8$

Step-by-step explanation:

The equation of the line in point slope form is

$y-y1=m(x-x1)$

we have

$y-1=(1/4)(x-12)$

so

$m=1/4$

$(12, 1)$

step 2

Find the equation of the line in standard form

The equation of the line in standard form is

$Ax+By=C$

where

A is positive integer

B and C are integers

we have

$y-1=(1/4)(x-12)$

Multiply by 4 both sides to remove the fraction

$4y-4=(x-12)$

$x-4y=12-4\\x-4y=8$

8 0

2 years ago

Read 2 more answers

The segments shown below could form a triangle.

Rina8888 [55]

Answer:

A. True

Step-by-step explanation:

The Triangle Inequality Theorem says that the sum of any two sides must be greater than the third side. Let's see if this is true.

a + b

9 + 1 = 10>9

a + c

9 + 9 = 18>1

c + b

9 + 1 = 10>9

5 0

2 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:

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