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Andreas93 [3]
3 years ago
5

Generally those who are less active have no opportunities to exercise

Mathematics
2 answers:
Alex777 [14]3 years ago
8 0
What is this question?
guajiro [1.7K]3 years ago
3 0

Answer:

False

Step-by-step explanation:

Edge 2021

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PLEASE HELP ME I HAVE A MATH TEST DUE TODAY AND THIS IS THE LAST QUESTION!!!! "The public library has volunteers shelve books af
liubo4ka [24]

Answer:

about 7  books per minute

Step-by-step explanation:

4 0
3 years ago
The Length of one side of a parallelogram is 3 more than twice the length of the adjacent side. The perimeter of the parallelogr
katovenus [111]

Answer: 4 cm and 11 cm.


Step-by-step explanation:

1. You can calculate the perimeter of a parallelogram as following:

P=a+b+c+d

Where a+b+c+d are the lenghts of the sides.

By definition the parallelogram has two pairs of parallel sides, therefore:

a=c and b=d

Then, you can calculate the perimeter as following:

 P=2(a+b)

2. Let's call the adjacent side x, then the other side is 2x+3.

3. Substitute values  and solve for x:

30=2(x+2x+3)\\30=2(3x+3)\\30=6x+6\\24=6x\\x=4

4. Then the lenghts of the sides are:

x=4cm\\2x+3=2(4cm)+3=11cm


5 0
4 years ago
Katrina drove 140 miles in two hours. If she continues driving in that speed, how long will it take her to drive 350 miles?
Vika [28.1K]

Answer:

210

Step-by-step explanation:


5 0
4 years ago
Help is required, I am stuck
Natalija [7]

Answer:

see explanation

Step-by-step explanation:

Given

x² + 4x - 7 = 0 ( add 7 to both sides )

x² + 4x = 7

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(2)x + 4 = 7 + 4

(x + 2)² = 11 ( take the square root of both sides )

x + 2 = ± \sqrt{11} ( subtract 2 from both sides )

x = - 2 ± \sqrt{11}

Thus

x = - 2 - \sqrt{11} ≈ - 5.32 ( to 2 dec. places )

x = - 2 + \sqrt{11} ≈ 1.32 ( to 2 dec. places )

8 0
3 years ago
How can you use the distributive property and mental math to find 5 x 198
Anarel [89]

Subtracte 198 then add what was taken off to 5

6 0
4 years ago
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