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JulijaS [17]
3 years ago
5

A building that is 150 m tall, 100 m wide, and 80 m deep has glass on all sides except the roof and the base.

Mathematics
1 answer:
Airida [17]3 years ago
3 0

Answer:

C. 54,000 m^2

Step-by-step explanation:

2 sides + 2 more sides

2(150x100) + 2(150x80)

30000 + 24000

54000m^2

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If (3). write the following vectors in column vector form:
Art [367]

Answer:

see explanation

Step-by-step explanation:

Given

a = \left[\begin{array}{ccc}3\\2\\\end{array}\right]

To obtain -3a multiply each of the elements of a by -3

3a = \left[\begin{array}{ccc}-3(3)\\-3(2)\\\end{array}\right] = \left[\begin{array}{ccc}-9\\-6\\\end{array}\right]

To obtain 1.5a multiply each element by 1.5

1.5a = \left[\begin{array}{ccc}1.5(3)\\1.5(2)\\\end{array}\right] = \left[\begin{array}{ccc}4.5\\3\\\end{array}\right]

8 0
3 years ago
Can you show the steps <br> thanks .........
Andrews [41]

Answer:

40 black bears

Step-by-step explanation:

Hope it helps.

5 per sample

8 samples

5 x 8 = 40 black bears

4 0
2 years ago
A) Serena walks 4 km in 1 h.<br><br> Explain!
Shkiper50 [21]

Answer:

didn't I explain how to do it

6 0
3 years ago
Find the area of this shape below
rjkz [21]

Answer: 28                      

Step-by-step explanation:

2 × 8 = 16

8 + 4 = 12

12 ÷ 2 = 6

6 × 2 = 12

16 + 12 = 28

6 0
3 years ago
4. Find the standard from of the equation of a hyperbola whose foci are (-1,2), (5,2) and its vertices are end points of the dia
Ad libitum [116K]

The <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

<h3>How to find the standard equation of a hyperbola</h3>

In this problem we must determine the equation of the hyperbola in its <em>standard</em> form from the coordinates of the foci and a <em>general</em> equation of a circle. Based on the location of the foci, we see that the axis of symmetry of the hyperbola is parallel to the x-axis. Besides, the center of the hyperbola is the midpoint of the line segment with the foci as endpoints:

(h, k) = 0.5 · (- 1, 2) + 0.5 · (5, 2)

(h, k) = (2, 2)

To determine whether it is possible that the vertices are endpoints of the diameter of the circle, we proceed to modify the <em>general</em> equation of the circle into its <em>standard</em> form.

If the vertices of the hyperbola are endpoints of the diameter of the circle, then the center of the circle must be the midpoint of the line segment. By algebra we find that:

x² + y² - 4 · x - 4 · y + 4= 0​

(x² - 4 · x + 4) + (y² - 4 · y + 4) = 4

(x - 2)² + (y - 2)² = 2²

The center of the circle is the midpoint of the line segment. Now we proceed to determine the vertices of the hyperbola:

V₁(x, y) = (0, 2), V₂(x, y) = (4, 2)

And the distance from the center to any of the vertices is 2 (<em>semi-major</em> distance, a) and the semi-minor distance is:

b = √(c² - a²)

b = √(3² - 2²)

b = √5

Therefore, the <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

To learn more on hyperbolae: brainly.com/question/27799190

#SPJ1

5 0
1 year ago
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