1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Pie
3 years ago
10

’ *5 Rec-tan ole 1 8 1' Rectangle 2 h - Rectangle 3 5 n . T - .22; - surface area of the figure ‘

Mathematics
1 answer:
yanalaym [24]3 years ago
8 0
I don't quite understand your topic sentence however to my best knowledge This is what i've come up with:
Rectangle 1 and 3 are the same. The only known side is the width at 5 inches whilst the other is unknown so it will be h.
Rectangle 1 and 3 area are "5h"

Rectangle 2 has a width of 8 and a length of h. Its area is "8h"

Without triangles our TSA for the rectangles is:   2 X 5H + 8H = TSA

The triangles are both assumed to be the same are. To get the Are of one it is \frac{1}{2}*(B*H)
Which = 12

Ans: TSA = 18H + 24

You might be interested in
A hypothetical square grows so that the length of it's diagonals are increasing at a rate of 8 m/min. How fast is the area of th
guapka [62]

Answer:

32m^{2}/min

Step-by-step explanation:

We let the length of the square be l, the diagonal be z and the area be A.

Then by Pythagoras theorem;

l^{2}+l^{2}=z^{2}\\2l^{2}=z^{2}

The are of a square of length l is given as;

A=l^{2}

Combining these two equations yields;

z^{2}=2A

A=\frac{z^{2} }{2}

Next, we have been given;

\frac{dz}{dt}=8

We are required to find;

\frac{dA}{dt} when z = 4

By chain rule;

\frac{dA}{dt}=\frac{dA}{dz}*\frac{dz}{dt}

Differentiating A=\frac{z^{2} }{2} with respect to z yields;

\frac{dA}{dz}=z

Therefore,

\frac{dA}{dt}=\frac{dA}{dz}*\frac{dz}{dt} = 8z

When z is 4m the area of the square will be increasing at a rate of;

8m/min * 4m = 32m^{2}/min

5 0
4 years ago
Pls help me answer this question and I help you answer a question
wel

Answer:

she can pick each marble and she can get two heads or two tails

Step-by-step explanation:

8

8 0
3 years ago
Read 2 more answers
Sophia and Tyler are running around a 400-meter track. They start running from the same place, at the same time. Sophia runs at
Blizzard [7]
To answer this item, we let x be the time in which Tyler and Sophia will be able to cover the same distance along the circular track. Since Sophia runs faster, she will be able to finish the track and y meters more while Tyler will be able to finish y meters only. 
                                             5(x)  = 400 + y
                                             4(x) = y
The values of x and y are 400 and 1600. Thus, it will take 400 seconds for then to meet again. 
8 0
3 years ago
Section 1
Gelneren [198K]

Answer:

The correct answer is part c:132

3 0
3 years ago
These cylinders are similar. Find the surface area of the smaller cylinder. Round to the nearest tenth.
Ratling [72]

Answer:  

Surface Area = 6.28 x 5.223 x 3 + 6.28 x 10.446 = 164 cm^2

Step-by-step explanation:

A=2πrh+2πr2

When we divide 245 by 1.5 we find 164cm^2

6 0
4 years ago
Other questions:
  • Akira receives a prize at a science fair for having the most informative project. Her trophy is in the shape of a square pyramid
    7·2 answers
  • For a line perpendicular to the line y = 2x - 3 through the point (4,-6)
    14·2 answers
  • 2cos^2x+sinx=2 domain restriction 0,2pi
    14·1 answer
  • Hellllllllllllllpppppppppp
    10·2 answers
  • Two painters can paint a room in 4 hours if they work together. The less experienced painter takes 3 hours more than the more ex
    8·1 answer
  • HELLOO PLEASE HELP MEEEE !!!
    15·1 answer
  • Can someone help me with this please
    13·2 answers
  • HELP WILL GIVE BRAINLEIST
    10·2 answers
  • What are the lengths of the legs of a right triangle in which one acute angle measures 19° and the hypotenuse is 15 units long?
    6·1 answer
  • What is the answer to x > 10
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!