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jeka94
2 years ago
13

Find the coordinates of the midpoint of1. (6,3) and (4,5)​

Mathematics
1 answer:
expeople1 [14]2 years ago
8 0

Answer:

(5,4)

Step-by-step explanation:

midpoint = (6 + 4 / 2 , 3 +5 / 2)

= (5,4)

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Step-by-step explanation:

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Solve the equation for a.<br> K=4a+9ab
Lena [83]

Answer:

a=  k/9b+4

Step-by-step explanation:

https://www.mathpapa.com/algebra-calculator.html?q=K%3D4a%2B9ab

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3 years ago
How much cardboard is needed to make the tissue box above? Square inches 8 3 3
Katen [24]

The amount of cardboard needed to make a cuboid box of dimensions 8 inch, 3 inches and 3 inches is 114 sq. inches.

<h3>What is the surface area of cuboid?</h3>

Let the three dimensions(height, length, width) be x, y,z units respectively.

The surface area of the cuboid is given by

S = 2(a\times b + b\times c + c\times a)

For this case, tissue box is almost cuboid shaped.

Also, its dimensions are given being 8 inches, 3 inches and 3 inches.

Suppose we measure the amount of cardboard needed in terms of area, then, the amount of cardboard needed to make that box(without any whole, full cuboid) is equal to the area of its surface(either outer or inner if we assume 0 inches thickness of cardboard),

Thus, we get:

Amount of cardboard needed = surface area of cuboid box with dimensions 8 by 3 by 3 (in inches)

= 2(8 \times 3 + 3 \times 3 + 3 \times 8) = 114 \: \rm in^2

Thus, the amount of cardboard needed to make a cuboid box of dimensions 8 inch, 3 inches and 3 inches is 114 sq. inches.

Learn more about surface area of cuboid here:

brainly.com/question/13522634

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5 0
2 years ago
Tutorial Exercise A boat leaves a dock at 2:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due
ANTONII [103]

Answer:

Both the boats will closet together at 2:21:36 pm.

Step-by-step explanation:

Given that - At 2 pm boat 1 leaves dock and heads south and boat 2 heads east towards the dock. Assume the dock is at origin (0,0).

Speed of boat 1 is 20 km/h so the position of boat 1 at any time (0,-20t),

        Formula :   d=v*t

at 2 pm boat 2 was 15 km due west of the dock because it took the boat 1 hour to reach there at 15 km/h, so the position of boat 2 at that time was (-15,0)

the position of boat 2 is changing towards east, so the position of boat 2 at any time (-15+15t,0)

      Formula : D=\sqrt{(x2-x1)^2+(y2-y1)^2}

⇒                     D = \sqrt{20^2t^2+15(t-1)^2}

Now let           F(t) = D^2(t)

                ∵    F'(t) = 800t + 450(t-1) =  1250t -450\\F'(t) =0

⇒                     t= 450/1250

⇒                     t= .36 hours

⇒                       = 21 min 36 sec

Since F"(t)=0,

∴ This time gives us a minimum.

Thus, The two boats will closet together at 2:21:36 pm.

3 0
3 years ago
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