What is what is the area of this face 10 inches 5 in 20 in 10 in 5 in 10

What is what is the area of this face 10 inches 5 in 20 in 10 in 5 in 10

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  1. Cirrhosis of the liver

    Unlawful speed resulting in a collision

    Have life-long disabilities

    No period of time

    Gender

    10%

    Under 40 inches

    The recommended driving speed for that area

    False

    Behind the crossing symbol painted on the road

    70 mph

    Both

    I would say false, but I’m not sure for this one

    .03

    The person might be in an emergency situation

    False

    A hospital zone

    20 inches

    .02

    New Year’s Eve

    .08

    A felony for drug possession

    True

    Water builds up on the road

    False

    $1,000

    Test your breaks when you get to the other side

    Under any circumstances

    Hope I helped!!

  2. Choice D)
    S(x) = 6x^2 - 20x

    Explanation:

    x = side length of base
    x*x = x^2 = area of base

    The top also has an area of x^2 since the base and top are both congruent squares. The total base area is x^2+x^2 = 2x^2

    The height h is 5 inches shorter than the base, so
    h = (base length) - 5
    h = x-5

    Each lateral side is of area h*x = (x-5)*x = x^2-5x

    There are 4 lateral sides
    Total lateral area = 4*(area of one lateral side)
    Total lateral area = 4*(x^2-5x)
    Total lateral area = 4*x^2-4*5x
    Total lateral area = 4*x^2-20x

    Add the total lateral area (4x^2-20x) to the total base area (2x^2)

    Doing so gets us
    S(x) = Total Surface Area
    S(x) = (Area of bases) + (area of lateral sides)
    S(x) = (2x^2) + (4x^2-20x)
    S(x) = (2x^2+4x^2) - 20x
    S(x) = 6x^2 - 20x
    which is why the answer is choice D

  3. The answer is the option C

    Net A with [tex]510[/tex] square inches

    Step-by-step explanation:

    we know that

    The surface area is equal to the sum of the areas of all the faces

    In this problem the correct Net is the figure A

    To find the surface area find the area of the [tex]3[/tex] rectangles plus the area of the [tex]2[/tex] triangles

    so

    Find the area of the [tex]3[/tex] rectangles

    [tex]A1=13*15+12*15+5*15=450\ in^{2}[/tex]

    Find the area of the [tex]2[/tex] triangles

    [tex]A2=12*5/2+12*5/2=60\ in^{2}[/tex]

    Adds the areas

    [tex]450\ in^{2}+60\ in^{2}=510\ in^{2}[/tex]

  4. As far as I can tell you are right on all but the last one and the second one because you didn't answer. The last one you got the area of one face but there are 6, and the second one is 30.52 because you have to get 3 1/8 as a decimal which is 3.125 and then you cube it or multiply it by itself then the product by 3.125 again.

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