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3 years ago

13

Shane is using pieces of square paper to cover a cube shaped gift with sides that are 9 inches long to find out the area of the

paper he needs he uses the formula for the surface area of cube A=6s2 What is the surface area of the cube

1 answer:

nordsb [41]3 years ago

3 0

Answer:

The area of the cube is 486 inches^2

Step-by-step explanation:

In this question, we are tasked with calculating the area of cube with side 9 inch.

Mathematically, the area can be calculated using the formula A = 6s^2

Now, what we need to do is to substitute 9 inches for s

Thus, A = 6 * 9^2

A = 6 * 81

A = 486 inches^2

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Answer:

13) 7 + x/2 = 10

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16) 6x - 6 = 12

17) x/3 + 10 = 12

18) 2x + 7 = 1

19) 9 + x/7 = 11

20) 8(n - 3)

Step-by-step explanation:

The quotient of x and 2 = x ÷ 2 = x/2

The product is the result of multiplying two or more other numbers

The sum is the result of adding two or more numbers

The difference is the result of subtracting one number from another

13) 7 + x/2 = 10

14) 2x - 5 = 7

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Answer:

The answer is -4x^2

Step-by-step explanation:

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Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

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$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
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Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
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Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
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We can conclude that the perimeter of our triangle is 140.13 square units.

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