Surface Area Gift Wrapping Drag and Drop Activity What is the answer for 2,5,6, & seven? I need it quick please !!

Answers

Answer 1

In this Drag and Drop Activity Answer 2nd is 162 [tex]in^{2}[/tex],  Answer 5th is 384 [tex]in^{2}[/tex],  Answer 6th is 1392 [tex]in^{2}[/tex] , Answer 7th is 1 roll.

2) In the 2nd figure, we have 5 faces as two equal Right angle Triangles (on two opposite faces) and three rectangles(on three sides of triangles)

Dimensions of both Right Angle Triangle:

Perpendicular = 9in

Base = 12 in ;

Dimensions of Rectangles:

Rectangle 1 : L1 = 15 in

                    B1 = 3 in

Rectangle 2: L2 = 9 in

                    B2 = 3 in

Rectangle 3: L3 = 12 in

                    B3 = 3 in

Area of 2nd figure = Sum of the area of its 5 faces

= Area of two Right angle Triangles + Area of Rectangle 1 + Area of rectangle 2 + Area of Rectangle 3

= [tex]\frac{1}{2}[/tex] * perpendicular * Base + L1*B1 + L2*B2 + L3*B3   ...................... (Formulas for particular Figures)

= [tex]\frac{1}{2}[/tex]* 9*12 + 15*3 + 9*3 + 12*3 [tex]in^{2}[/tex]

= 54+ 45+ 27 + 36 [tex]in^{2}[/tex]

=162 [tex]in^{2}[/tex]

Area of 2nd figure = 162 [tex]in^{2}[/tex]

5) In the 5th figure, we have a cube with all sides equals to 8in. Cube has 6 faces.

The area of the figure is = [tex]6(side)^{2}[/tex]

= [tex]6(8)^{2}[/tex][tex]in^{2}[/tex]

=384 [tex]in^{2}[/tex]

Area of 5th figure = 384 [tex]in^{2}[/tex]

6) Total Amount of wrapping paper needed is the sum of the Area of all Figures

= 364 + 162 + 290 + 192+ 384 [tex]in^{2}[/tex]

= 1392 [tex]in^{2}[/tex]

The total amount of wrapping paper needed is = 1392 [tex]in^{2}[/tex]

7) One Roll of wrapping paper with measurements 30 in and 60 in has an Area

= 30*60 [tex]in^{2}[/tex]

= 1800 [tex]in^{2}[/tex]

As we need only 1392 [tex]in^{2}[/tex] of wrapping paper which is less than one roll of wrapping paper so only 1 roll is required

hence, In this Drag and Drop Activity Answer 2nd is 162 [tex]in^{2}[/tex],  Answer 5th is 384 [tex]in^{2}[/tex],  Answer 6th is 1392 [tex]in^{2}[/tex] , Answer 7th is 1 roll.

To learn more about Area of Rectangle

https://brainly.com/question/2607596

Answer 2
2nd because of the dimensions of the triangle

Related Questions

Sanderson is having trouble with his assignment. His shown work is as follows:− 8/9 div 9/8 = − 89 × 98 = − 72/72 = −1 v\However, his answer does not match the answer that his teacher gives him.


Complete the description of Sanderson's mistake. Find the correct answer.

He forgot to multiply by the
(select)
, and instead just multiplied by the
(select)
that was in the denominator. The answer should look like this:
− 8/9 div 9/8 = - 8/9 × 8/9 =

need this for final exams

Answers

Answer:

He forgot to flip the second fraction.

Step-by-step explanation:

He has:

−8/9 ÷ 9/8 = −8/9 × 9/8 = −72/72 = −1

He made a mistake. When you change a division to a multiplication, you must use the reciprocal of the second fraction. You have to flipt the second fraction. He did not flip it.

This is correct:

−8/9 ÷ 9/8 = −8/9 × 8/9 = −64/81

He just didn’t flip the second fraction, he had to use the reciprocal

Please help, screenshot!

Answers

Answer:

4/7

Step-by-step explanation:

class total = 21

chance of picking one of 12 boys out of 21.

12/21 which can be simpilfied to 4/7

4/7

9 girls + 12 boys = 21
Boys/total = 12/21
12/21 divided by 3 = 4/7

HELP ASAP!! A student is painting a brick for his teacher to use as a doorstop in the classroom. He is only painting the front of the brick. The vertices of the face are (−4, 2), (−4, −11), (4, 2), and (4, −11). What is the area, in square inches, of the painted face of the brick?
A. 42 in2
B. 52 in2
C. 104 in2
D. 128 in2

Answers

Answer:

104in2

Step-by-step explanation:

The area, in square inches, of the painted face of the brick is Option C 104 [tex]in^{2}[/tex].

Using the Quadrant system of the Number line

B (-4,2) lies in the 2nd quadrant.

C (-4,-11) lies in the 3rd quadrant

A (4,2) lies in the 1st quadrant

D (4,-11) lies in the 4th quadrant

Length of Side AB= Length of Side CD= Distance between x coordinate of A & B OR Between C & D i.e. 8 in

Length of Side BC= Length of side  AD= Distance between y coordinate of A & D OR Between B & C i.e. 13 in

Area of the Square ABCD= (length of Side AB or CD) X (Length of Side  BC or AD)

                                          = (8 x 13) [tex]in^{2}[/tex]

                                          = 104 [tex]in^{2}[/tex]

Thus, Area of square is 104 [tex]in^{2}[/tex].

To learn more about area: https://brainly.com/question/25292087

A chemist wants to dilute a 30% phosphoric acid solution to a 15% solution. He needs 10 liters of the 15% solution. How many liters of the 30% solution and water must the chemist use?

Answers

The chemist needs to use 5 liters of the 30% solution and 5 liters of water.

Answer:

30% solution x liters

30%(water) 10-x liters

15% 10 liters

Step-by-step explanation:

the sum of the reciprocals of two consecutive even intergers is 9/40 this can be represented bby the equation shown 1/x+1/x+2=9/40 use the rational equation to determine the integers
SHOW YOUR WORK PLEASE!!!!!!!

Answers

Answer:

In explanation

Step-by-step explanation:

To solve the rational equation 1/x + 1/(x + 2) = 9/40, we can start by finding a common denominator and then simplifying the equation. The common denominator for the two fractions is (x)(x + 2). Multiplying each term by this common denominator, we get:

[(x + 2) + x] / (x)(x + 2) = 9/40

Simplifying the numerator, we have:

(2x + 2) / (x)(x + 2) = 9/40

Now, we can cross-multiply to eliminate the denominators:

40(2x + 2) = 9(x)(x + 2)

Expanding both sides, we have:

80x + 80 = 9x^2 + 18x

Rearranging the equation and setting it equal to zero:

9x^2 - 62x - 80 = 0

Now, we can factor the quadratic equation:

(9x + 10)(x - 8) = 0

Setting each factor equal to zero, we have:

9x + 10 = 0 or x - 8 = 0

Solving for x in each case:

9x = -10 or x = 8

Dividing both sides of the first equation by 9, we find:

x = -10/9 or x = 8

Since we are looking for even integers, we can disregard the negative solution. Therefore, the value of x is 8.

Hence, the two consecutive even integers can be represented by x and x + 2, which gives us 8 and 10. So, the integers are 8 and 10.

Its 7840 bc id you see how you have it put together you just make it all ip

Keira said that if 6 tickets are sold, there should be 2 tickets left. Is Keira correct?

Answers

To determine whether Keira's statement is correct, we need to analyze the information provided.

According to Keira, if 6 tickets are sold, there should be 2 tickets left. However, we don't have any information about the initial number of tickets available.

If the total number of tickets initially was 8, and 6 tickets were sold, then there would indeed be 2 tickets left, supporting Keira's statement.

However, if the total number of tickets initially was less than 8, or if the statement doesn't specify the initial number of tickets, we cannot determine if Keira's statement is correct.

Therefore, without additional information, we cannot determine the accuracy of Keira's statement.
in order to know if Keira has two tickets left, we need to know the first number of tickets without that number there is no way to find the remainder

Please help! -screenshot-

Answers

25% I think for second one!
25% pretty sure on second ine

Can someone help me pls

Answers

for problem one, lets say
X = 2
Y = 6
z = 1
( these are just for example)
x ( y > z ) = y ( x > z )
they both have the same answers so they are equivalent


now for problem 2.. i dont know the answer but theres two of the ( p v q ) and yeah i dont know how to explain this one im sorry
For Problem 2, you can see its law of syllogism
which is:
p -> q,
q -> r
p -> r

p v q = p OR q,
p n q = p AND q (n is the v upside down)

so you need to give an example of how

p -> (q AND r) can be equal to (NOT p, OR q) and (NOT p, OR r)

just find 3 expressions or sentences that can plug into p q and r, and there you have it :)

15 points to whoever gets this right <3

Answers

Answer:

77.5

Step-by-step explanation:

Given:

Scores: 65, 70, 75, 80, 85, 90, 95

Number of Students: 8, 5, 3, 6, 6, 4, 2

Step 1: Calculate the sum

Sum = (65 * 8) + (70 * 5) + (75 * 3) + (80 * 6) + (85 * 6) + (90 * 4) + (95 * 2)

= 520 + 350 + 225 + 480 + 510 + 360 + 190

= 2635

Step 2: Calculate the mean

Mean = Sum / Total Number of Students

= 2635 / (8 + 5 + 3 + 6 + 6 + 4 + 2)

= 2635 / 34

≈ 77.5

Answer:77.5

Step-by-step explanation:

Does the series converge or diverge? if it converges, what is the sum? SHOW YOUR WORK!!!!!!
equation is in picture.

Answers

Answer:  It converges to   -3

=============================================

Reason:

This is a geometric series with

a = -4 = first termr = -1/3 = common ratio

The template is [tex]a(r)^{n-1}[/tex]

If -1 < r < 1, then the infinite geometric series converges to a finite number. This is because we add on smaller and smaller pieces, which prevents the sum going off to infinity.

In the case of r = -1/3, it fits the interval  -1 < r < 1. In other words -1 < -1/3 < 1 is true.

We'll plug those values into the formula below to wrap things up.

[tex]S = \frac{a}{1-r}\\\\S = \frac{-4}{1-(-1/3)}\\\\S = \frac{-4}{1+1/3}\\\\S = \frac{-4}{3/3+1/3}\\\\[/tex]

[tex]S = \frac{-4}{4/3}\\\\S = -4 \div \frac{4}{3}\\\\S = \frac{-4}{1} \times \frac{3}{4}\\\\S = \frac{-4*3}{1*4}\\\\S = -3\\\\[/tex]

Therefore,

[tex]\displaystyle \sum_{n=1}^{\infty} -4\left(-\frac{1}{3}\right)^{n-1} = -3[/tex]

The final answer is -3.

You can verify the answer by generating partial sums with a spreadsheet. The partial sums should steadily get closer to -3.

Here's a few partial sums.

[tex]\begin{array}{|c|c|c|} \cline{1-3}\text{n} & \text{a}_{\text{n}} & \text{S}_{\text{n}}\\\cline{1-3}1 & -4 & -4\\\cline{1-3}2 & 1.333333 & -2.666667\\\cline{1-3}3 & -0.444444 & -3.111111\\\cline{1-3}4 & 0.148148 & -2.962963\\\cline{1-3}5 & -0.049383 & -3.012346\\\cline{1-3}6 & 0.016461 & -2.995885\\\cline{1-3}7 & -0.005487 & -3.001372\\\cline{1-3}8 & 0.001829 & -2.999543\\\cline{1-3}9 & -0.00061 & -3.000153\\\cline{1-3}10 & 0.000203 & -2.99995\\\cline{1-3}\end{array}[/tex]

The interesting thing is that the partial sums [tex]S_n[/tex]  bounce around -3 while also getting closer to it.

Yes the partial sums should get closer to -3.
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