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3 years ago

7

Daniel brought 2.75 pounds of Spanish fried rice to a family dinner. During the meal, Daniel and his family ate 1.25 pounds of r

ice. At the end, Daniel's friend took home half of the rice that was left after dinner.

1 answer:

pickupchik [31]3 years ago

6 0

There was .75 left of Spanish fried rice.

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{HIGH PAYOUT} Can someone help by showing work on how to get to answer?

Gala2k [10]

Let's assume each line is one toothpick.

In figure 1, there is a square made out of the 4 toothpicks.

In figure 2, there are four squares made. However, the 4 four toothpicks were already used, so we don't count them twice. That gives 4 (that were used) + 3 (the new ones on the left) + 3 (new ones on top) + 3 (new ones on right). We sum them up: 4 + 3 + 3 + 3 = 13.

In figure 3 there are nine squares made. We have the 13 squares made, plus 3 on the left column, 2 on the left center column, 3, on the center column, 2 on the right center, and 3 on the right. We total them up. 13 + 3 + 2 + 3 + 2+ 3 = 26.

Let's look at the pattern.

<u>Figure #</u> <u># Toothpick</u>s

1 4

2 13

3 26

From figure 1 to 2, we added 9. From 2 to 3 we added 13. If we draw figure 4, we would add (from left to right) 3, 2, 2, 3, 2, 2, 3. That would be 17 toothpicks added, and 39 total. Let's go back to the table.

<u>Figure #</u> <u># Toothpick</u>

1 4

2 13

3 26

4 39

Now let's look at the sequence happening. As we add more, we increase by 9, then 13, then 17. For each row, we add the same number as the previous row and then add 4. Drawing Figure 5 would be too much, but we can conclude based on the pattern that 21 squares would be added. We can take this sequence as far as we need to. We want the 10th figure or term, so we apply the pattern and keep going until we get there.

<u>Figure #</u> <u># Toothpick</u>

1 4

2 13 <--- add 9

3 26 <--- add 13, the same 9 and 4 more

4 39 <--- add 17, the same 13 and 4 more

5 60 <--- add 21, the same 17 and 4 more

6 85 <-- add 25, the same 21 and 4 more

7 114 <-- add 29, the same 25 and 4 more

8 147 <-- add 33

9 184 <---- add 37

10 225 <-- add 41

Thus the 10th figure has 225 toothpicks.

5 0

3 years ago

A fair spinner has 9 equal sections: 4 red, 3 blue and 2 green.

love history [14]

Answer:

Probability would be 8/18 or 4.44%

Step-by-step explanation:

If its spun twice it acts as if it has 18 sections meaning you multiply the number of reds and number of squares total by two in this situation and put them over each other!

4 0

3 years ago

The sum of a number and 12

Blizzard [7]

Answer:

D) n+12

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a

Rashid [163]

Answer:

Length of original rectangle: 11 units.

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

$\text{Perimeter of new rectangle}=20$

Step-by-step explanation:

Let x represent width of the original rectangle.

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be $x+6$ .

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be $x-2$ .

The length of new rectangle would be $x+6-4=x+2$ .

The area of new rectangle would be $(x+2)(x-2)$ .

Now we will equate area of new rectangle with 21 and solve for x as:

$(x+2)(x-2)=21$

Applying difference of squares, we will get:

$x^2-2^2=21$

$x^2-4=21$

$x^2-4+4=21+4$

$x^2=25$

Since width cannot be negative, so we will take positive square root of both sides.

$\sqrt{x^2}=\sqrt{25}$

$x=5$

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be $x+6\Rightarrow x+5=11$ .

Therefore, the length of original rectangle is 11 units.

$\text{Area of original rectangle}=5\times 11$

$\text{Area of original rectangle}=55$

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

We know that perimeter of rectangle is two times the sum of length and width.

$\text{Perimeter of new rectangle}=2((x+2)+(x-2))$

$\text{Perimeter of new rectangle}=2((5+2)+(5-2))$

$\text{Perimeter of new rectangle}=2(7+3)$

$\text{Perimeter of new rectangle}=2(10)$

$\text{Perimeter of new rectangle}=20$

Therefore, the perimeter of the new rectangle is 20 units.

7 0

4 years ago

Lauren bought a rectangular box that is 6 inches long, 10 inches wide and 4 inches high. she wants to cover it in fabric. the fa

kirza4 [7]

First you'll need to find the area of each side and then add those areas together. 2(6x10)+2(4x6)+2(4x10)= 120+48+80=248
then you multiply the total area by .25 and get 62. It will cost $62.

3 0

3 years ago