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Tatiana [17]
3 years ago
5

Does anyone have the rest of the test? Surface Area and Volume Unit Test????

Mathematics
1 answer:
jarptica [38.1K]3 years ago
8 0

Answer:

B) 112 cm²; 336 cm²

Step-by-step explanation:

The lateral area would be without the bases. In this case, the bases are the top and the bottom

Lateral Area

(2)(2.54)(8) = 40.64

(2)(2.54)(14) = 71.12

Add together and get 111.76 cm²

Surface Area

LA + Bases

<em>Bases</em>

(2)(14)(8) = 224

Add with lateral area and get 335.76 cm²

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In one year, Linda's parents spend $2400 on cable and internet service. If they spend the same amount each month, what is the re
Yuri [45]

Answer: $200

Step-by-step explanation:

Given: In 1 year ,

Linda's parents spend $2400 on cable and internet service.

If they spend the same amount each month, Monthly payment = \dfrac{\text{Total money spent in 1 year}}{\text{Number of months in a year}}

=\dfrac{\$2400}{12}\ \ \ [\text{Number of months in a year}=12]\\\\=\$200

Hence, the resulting monthly change in the family's income = $200

4 0
3 years ago
- Two consecutive integers are 5 and 6. Write a quadratic equation that
mario62 [17]

Answer:

  x^2 -11x +30 = 0

Step-by-step explanation:

If these two integers are solutions of the quadratic, then its factors are ...

  (x -5)(x -6) = 0

Multiplying this out, we get ...

  x^2 -11x +30 = 0

6 0
3 years ago
What is the area of the trapezoid?
EastWind [94]

Answer:

37.5 square units

Step-by-step explanation:

A=a+b/2h => 5 + 10 / 2 x 5

7 0
2 years ago
Read 2 more answers
I'm have some issues with the problem shown in the screenshot and would love some help.
KiRa [710]

The dimensions and volume of the largest box formed by the 18 in. by 35 in. cardboard are;

  • Width ≈ 8.89 in., length ≈ 24.89 in., height ≈ 4.55 in.

  • Maximum volume of the box is approximately 1048.6 in.³

<h3>How can the dimensions and volume of the box be calculated?</h3>

The given dimensions of the cardboard are;

Width = 18 inches

Length = 35 inches

Let <em>x </em>represent the side lengths of the cut squares, we have;

Width of the box formed = 18 - 2•x

Length of the box = 35 - 2•x

Height of the box = x

Volume, <em>V</em>, of the box is therefore;

V = (18 - 2•x) × (35 - 2•x) × x = 4•x³ - 106•x² + 630•x

By differentiation, at the extreme locations, we have;

\frac{d V }{dx}  =  \frac{d( 4 \cdot \:  {x}^{3}  - 106 \cdot \:  {x}^{2}  + 630\cdot \:  {x} )  }{dx} = 0

Which gives;

\frac{d V }{dx}  =12\cdot \:  {x}^{2}  -  212\cdot \:  {x} + 630 = 0

6•x² - 106•x + 315 = 0

x  =  \frac{ - 6 \pm \sqrt{106 ^2 - 4 \times 6 \times 315} }{2 \times 6}

Therefore;

x ≈ 4.55, or x ≈ -5.55

When x ≈ 4.55, we have;

V = 4•x³ - 106•x² + 630•x

Which gives;

V ≈ 1048.6

When x ≈ -5.55, we have;

V ≈ -7450.8

The dimensions of the box that gives the maximum volume are therefore;

  • Width ≈ 18 - 2×4.55 in. = 8.89 in.

  • Length of the box ≈ 35 - 2×4.55 in. = 24.89 in.

  • Height = x ≈ 4.55 in.

  • The maximum volume of the box, <em>V </em><em> </em>≈ 1048.6 in.³

Learn more about differentiation and integration here:

brainly.com/question/13058734

#SPJ1

7 0
1 year ago
PLEASE HELP ME with this question
vovangra [49]

Answer:

abc are correct

Step-by-step explanation:

i took the test

8 0
2 years ago
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