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Digiron [165]
2 years ago
6

Pls help I will give you 15 points only

Mathematics
1 answer:
QveST [7]2 years ago
6 0
N= 1/8 I’m pretty sure I did that last year
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The area of a trapezoid is 39 square millimeters. The height of the trapezoid is 6 millimeters. One of the base lengths of the t
krok68 [10]
39×2÷6=13
13-5=8 mm. The other base of the trapezoid is 8 mm. Let check it:
1/2(8+5)×6
=1/2×13×6
=39 square mm. Hope it help!
5 0
3 years ago
What is the sum of the following numbers. -7,5,-3,6,-4,7,-5,3-1,4​
Sav [38]

Answer:

Step-by-step explanation:

Tổng : -7,5+(-3,6)+(-4,7)+(-5,3)+(-1,4)=-22,5

3 0
2 years ago
Can someone help me with the ones that aren’t done
garri49 [273]

5:

C = 5/9(F - 32)

C = 5/9(F) - 17.777

-5/9(F) + C = -17.777

-5/9(F) = -C - 17.777

F = 5/9(C) + 32.0306

7 0
2 years ago
Read 2 more answers
Para encher 2/5 de uma piscina são necessários 60.000 litros de água. Qual é a capacidade dessa piscina?
natali 33 [55]

Answer:

A capacidade dessa piscina é de 150.000 litros de água.

Step-by-step explanation:

A capacidade da piscina é de x.

Para encher 2/5 de uma piscina são necessários 60.000 litros de água.

Isto implica que:

\frac{2x}{5} = 60000

2x = 60000*5

x = \frac{60000*5}{2}

x = 150000

A capacidade dessa piscina é de 150.000 litros de água.

8 0
2 years ago
Two cables support a 800​-lb ​weight, as shown. Find the tension in each cable.
jeka94

Answer:

  • 892 lb (right)
  • 653 lb (left)

Step-by-step explanation:

The weight is in equilibrium, so the net force on it is zero. If R and L represent the tensions in the Right and Left cables, respectively ...

  Rcos(45°) +Lcos(75°) = 800

  Rsin(45°) -Lsin(75°) = 0

Solving these equations by Cramer's Rule, we get ...

  R = 800sin(75°)/(cos(75°)sin(45°) +cos(45°)sin(75°))

     = 800sin(75°)/sin(120°) ≈ 892 . . . pounds

  L = 800sin(45°)/sin(120°) ≈ 653 . . . pounds

The tension in the right cable is about 892 pounds; about 653 pounds in the left cable.

_____

This suggests a really simple generic solution. For angle α on the right and β on the left and weight w, the tensions (right, left) are ...

  (right, left) = w/sin(α+β)×(sin(β), sin(α))

5 0
2 years ago
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