Could somebody please answer these four questions with simple yet explanatory answers?
Could somebody please answer these four questions with simple yet - 1

Answers

Answer 1
Answer: 1) Surface Area = 2(lw + lh + wh)
2(4*5+4*9+5*9) =202in^2
Surface Area=202in^2
Lateral Area= Perimeter of base * height
5+5=10  <--width+width
4+4=8 <---length + length 
10+8=18  <--total perimeter
18 * 9=162in^2  <--multiplied the height +perimeter of base  
Lateral Area=162in^2

2) Same concept as the previous one
 Surface Area = 2(lw + lh + wh)
2(4*2+4*5+5*2) =76in^2
Surface Area= 76in^2

Lateral Area
Lateral Area= Perimeter of base * height
4+4=8
2+2=4
8+4=12
12 x 5=60
Lateral Area= 60in^2  
Answer 2
Answer: Pooja answered #1 and #2 correct, so I shall make my life easier by answering #3 and #4 :P

#3 tests the concept of surface area. The gift wrap will be covering only the surface of each side. Surface area is all of the outside. Volume is all of the stuff inside.

So the surface area of rectangular prism can be found by the formula:
SA = 2(LW + LH + WH)
SA = 2(9*7 + 7*4 + 9*4)
SA = 2(63 + 28 + 36)
SA = 2(127)
SA = 254 in
²

#4 also needs you to find the surface area. 
So the surface area of a triangular prism can be found by this formula:
SA = bh + (\sf s_1 + s_2 + s_3)h
SA = (6.3*2) + (6.3 + 2.5 + 5.2)3
SA = 12.6 + (14*3)
SA = 12.6 + 42
SA = 54.6 cm²

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Help with #2 pls and show me how with work

Answers

remember
you can do ANYTHING to an equation as long as you do it to BOTH SIDES
also
a/a=1
ab/ab=1

so
we want to solve for h or 1h

T=pirh
divide both sides by pir (since pir/pir=1)
T/(pir)=1h
T/(pir)=h


C
T=πrh
πrh=T
(πr) * h=T
h=T/πr

Answer: C    h=T/πr

How can you use the Pythagorean Theorem to solve real-world problems?You can use the Pythagorean Theorem to find missing ________ (angles or lengths) in objects that are_________. ( squares, right triangles, or equilateral Triangles)

Answers

You can use the Pythagorean Theorem to find missing lengths in objects that are right triangles.

What is Pythagorean Theorem?

The Pythagorean Theorem is a mathematical principle that relates to right triangles. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

You can use the Pythagorean Theorem to find missing lengths in objects that are right triangles.

The theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

By rearranging the formula, you can solve for the missing length when given the lengths of the other two sides.

This can be applied to real-world problems involving measurements of objects or distances, such as determining the length of a ladder needed to reach a certain height on a wall or calculating the distance between two points on a map.

Learn more about Pythagorean Theorem here:

brainly.com/question/14930619

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1st answer is lengths and 2nd answer is right triangles

Multiply 0.023 and 10^2

Answers

0.023 x 10^2

0.023 x 100

2.3



your answer would be 2.3


Find the percent of the number. 11.3% of 700



A.
7.91


B.
79.1


C.
791


D.
7910

Answers

First you need to convert 11.3% from percentage to number. To do this you just simply have to move the decimal place two places to the left, which gives us .113. To find out what 11.3% of 700 is you would just multiply .113 by 700. 

.113 x 700 = 79.1

Answer is B.
700*0.113
=79.1 %

Answer is B.

A veterinarian assistant has a 20 pound bag of cat food if he feeds each cat 2/5 pounds how many cats can he feed

Answers

Answer:

50

Step-by-step explanation:

20 / (2/5)

To divide by a fraction, multiply by the reciprocal.

20 × (5/2)

50

He can feed 50 cats.

50 cats

20 pounds is 100/5 pounds when written with a denominator of 5. Then you divide 100 by 2 and get 50.

Solve the following inequality. |3n-2|-2<1

Answers

Answer: (-1)/(3) < n < (5)/(3)

Step-by-step explanation:

| 3n - 2 | - 2 < 1

              +2 +2

| 3n - 2 |       < 3

3n - 2 < 3     and     3n - 2 > -3

     +2 +2                     +2   +2

3n       < 5      and     3n      > -1

       n <  (5)/(3)       and         n > (-1)/(3)

(-1)/(3) < n < (5)/(3)

Interval Notation:  ((-1)/(3),(5)/(3))

Graph:  (-1)/(3) o--------------------o (5)/(3)


|3n-2|-2<1\n\n|3n-2|<3\n\n3n-2<3 \wedge 3n-2>-3\n\n3n<5 \wedge 3n>-1\n\nn<(5)/(3) \wedge n>-(1)/(3)\n\nn\in \left(-(1)/(3),(5)/(3)\right)