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KIM [24]
3 years ago
15

In the diagram below bd is parallel to xy what is the value of x

Mathematics
2 answers:
Dominik [7]3 years ago
7 0

Answer:

the answer is 76

Step-by-step explanation:

lara [203]3 years ago
7 0

It’s 83

Step-by-step explanation:

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2- Describe the cross section
kicyunya [14]

The cross-section cut by the plane yields a quadrilateral with length and width equal to the length and width of the face that is parallel to the plane.

4 0
3 years ago
Find the missing side length 12 cm 1,200 cm 20 cm ____ cm​
Korvikt [17]

Answer:

  • Missing side length is 5 cm.

Step-by-step explanation:

<u>Multiply 20 by 12:</u>

  • 20 * 12 = 240.

<u>Multiply 240 by 5:</u>

  • 240 * 5 = 1200.

<u>Multiply the lengths:</u>

  • 20 * 12
  • = 240 * 5
  • = 1200.
3 0
3 years ago
Lowest common factor of 72 and 108​
melamori03 [73]

Answer:

The lowest common factor of 72 and 108 is 216

5 0
3 years ago
Read the passage.
alexandr1967 [171]

Answer:

D sentence 5

Step-by-step explanation:

6 0
4 years ago
In 4 ​minutes, a conveyor belt moves 800 pounds of recyclable aluminum from the delivery truck to a storage area. A smaller belt
kkurt [141]

Answer:

It takes 2 minutes 4 seconds to move the cans to storage area if both the belts are used.

Step-by-step explanation:

Given - In 4 ​minutes, a conveyor belt moves 800 pounds of recyclable

            aluminum from the delivery truck to a storage area. A smaller belt

            moves the same quantity of cans the same distance in 6 minutes.

To find - If both belts are​ used, find how long it takes to move the cans

              to the storage area.

Proof -

Given that,

Larger belt take 4 minutes to transfer the aluminium from delivery truck to storage area.

⇒ Time taken by Larger belt = 4 min/ shifting

Also,

Smaller belt take 6 minutes to transfer the aluminium from delivery truck to storage area.

⇒ Time taken by Smaller belt = 6 min/ shifting

Now,

As we know that Time is inversely proportional to time, so

Rate of Larger Belt = \frac{1}{4} shifting/min

Rate of Smaller Belt = \frac{1}{6} shifting/min

Now,

Let us assume that , if both belts used then,

Total time taken = x min/shifting

Then, Rate = \frac{1}{x} shifting/min

Now,

\frac{1}{4}  + \frac{1}{6}  = \frac{1}{x}

⇒\frac{6 + 4}{24} = \frac{1}{x}

⇒\frac{10}{24} = \frac{1}{x}

⇒10x = 24

⇒x = \frac{24}{10} = 2.4 minutes

∴ we get

It takes 2 minutes 4 seconds to move the cans to storage area if both the belts are used.

3 0
3 years ago
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