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aleksandr82 [10.1K]
2 years ago
12

What fastens the gauge line at the end fittings on the end of a gauge line?

Mathematics
2 answers:
neonofarm [45]2 years ago
7 0
<span>Tighten the screw clamp over the fitting with a screwdriver, and place another metal screw clamp over the other end of the rubber vacuum line.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
</span>
Alla [95]2 years ago
4 0

Answer:

Step-by-step explanation:

Clip the screw clamp over the fitting at one end by making use of a screwdriver, set another metal screw, clamp it on top at the extreme end of the rubber vacuum line.

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Tickets were sold at four different gates of a high school football stadium. The graph below shows the percent of the total tick
MaRussiya [10]
you have to add 90 +90 3 times then theres your answer :)



6 0
2 years ago
joseph says that there is 5/8 of a ham sandwich left and 1/2 of a turkey sandwich left. what is NOT a pair of equivalent fractio
ra1l [238]

Two <em>possible answers</em> are:

5/16 and 3/8.

Explanation:

To create a fraction equivalent to a given fraction, you multiply the numerator and denominator by the same number. This means for 5/8, we could have:

5/8; (5*2)/(8*2)=10/16; (5*3)/(8*3)=15/24; etc.

For 1/2, we could have:

1/2; (1*2)/(2*2)=2/4; (1*3)/(2*3)=3/6; etc.

5 0
2 years ago
Read 2 more answers
Two friends went fishing on a lake. One friend's lure went 23 feet below the lake's surface, while the other friend's lure sank
serg [7]

Answer:

Distance = 58\ feet

Step-by-step explanation:

Represent the lure's difference with A and B;

A = 23\ feet

B = 81\ feet

Required

Determine the difference in depth between the lure's depth

The distance is calculated as follows;

Distance = B - A

Substitute 23 feet for A and 81 feet for B

Distance = 81\ feet - 23\ feet

Distance = 58\ feet

<em>Hence, the distance between both lure's is 58 feet</em>

6 0
2 years ago
Peter wants to cut a rectangle of size 6x7 into squares with integer sides. What is the smallest number of squares he can get
Semmy [17]
 He can cut 5 squares, by making 1 4 × 4, 2 3 × 3 and 2 2 × 2 squares.
5 0
2 years ago
Find the dimensions of a rectangle with area 512 m2 whose perimeter is as small as possible. (If both values are the same number
Masja [62]

Answer:

<h2>√512 by √512 </h2>

Step-by-step explanation:

Length the length and breadth of the rectangle be x and y.

Area of the rectangle A = Length * breadth

Perimeter P = 2(Length + Breadth)

A = xy and P = 2(x+y)

If the area of the rectangle is 512m², then 512 = xy

x = 512/y

Substituting x = 512/y into the formula for calculating the perimeter;

P = 2(512/y + y)

P = 1024/y + 2y

To get the value of y, we will set dP/dy to zero and solve.

dP/dy = -1024y⁻² + 2

-1024y⁻² + 2 = 0

-1024y⁻² = -2

512y⁻² = 1

y⁻² = 1/512

1/y² = 1/512

y²  = 512

y = √512 m

On testing for minimum, we must know that the perimeter is at the minimum when y = √512

From xy = 512

x(√512) = 512

x = 512/√512

On rationalizing, x = 512/√512 * √512 /√512

x = 512√512 /512

x = √512 m

Hence, the dimensions of a rectangle is √512 m  by √512 m

5 0
2 years ago
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