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
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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
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
Keira said that if 6 tickets are sold, there should be 2 tickets left. Is Keira correct?
Please help, screenshot!
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
Please help! -screenshot-
Does the series converge or diverge? if it converges, what is the sum? SHOW YOUR WORK!!!!!!
equation is in picture.
=============================================
Reason:
This is a geometric series with
a = -4 = first termr = -1/3 = common ratioThe 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.
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
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
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!!!!!!!
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.
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?
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:
15 points to whoever gets this right <3
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:
Can someone help me pls