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Vitek1552 [10]
3 years ago
5

YOU KNOW THE DRILL 2.0

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
6 0

Answer:

#1

Step-by-step explanation:

The four yellow boxes represent x so together they are 4 * x or 4x. The blue boxes seem to represent -1 and since there are three of them together they are -1 * 3 = -3. 4x + (-3) = 4x - 3.

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In answering the following questions, assume that the angular acceleration is constant and nonzero: alpha not equal to zero.
Ronch [10]
I think the correct answer would be true. The quantity represented by theta is a function of time. This is because the rate of change of angular speed is constant which means the angular distance is changing with time. Hope this answers the question. Have a nice day.
8 0
3 years ago
The coordinates of each rectangle ABCD are A (0,2) B (2,4) C (3,3) D (1,1). Find the distance of each side of the rectangle then
EastWind [94]

AB = CD = √8 ≈ 2.8 units

BC = AD = √2 ≈ 1.4 units

Area of the rectangle ABCD = 3.92 units²

Perimeter of the rectangle ABCD = 8.4 units

<h3>How to Find the Area and Perimeter of a Rectangle?</h3>

Given the coordinates of vertices of rectangle ABCD as:

  • A(0,2)
  • B(2,4)
  • C(3,3)
  • D(1,1)

To find the area and perimeter, use the distance formula to find the distance between A and B, and B and C.

Using the distance formula, we have the following:

AB = √[(2−0)² + (4−2)²]

AB = √[(2)² + (2)²]

AB = √8 ≈ 2.8 units

CD = √8 ≈ 2.8 units

BC = √[(2−3)² + (4−3)²]

BC = √[(−1)² + (1)²]

BC = √2 ≈ 1.4 units

AD = √2 ≈ 1.4 units

Area of the rectangle ABCD = (AB)(BC) = (2.8)(1.4) = 3.92 units²

Perimeter of the rectangle ABCD = 2(AB + BC) = 2(2.8 + 1.4) = 8.4 units

Learn more about the area and perimeter of rectangle on:

brainly.com/question/24571594

#SPJ1

5 0
1 year ago
If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?
Alisiya [41]
If k is odd, then

\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor

while if k is even, then the sum would be

\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4

The latter case is easier to solve:

\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0

which means k=6.

In the odd case, instead of considering the above equation we can consider the partial sums. If k is odd, then the sum of the even integers between 1 and k would be

S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)

Now consider the partial sum up to the second-to-last term,

S^*=2+4+6+\cdots+(k-5)+(k-3)

Subtracting this from the previous partial sum, we have

S-S^*=k-1

We're given that the sums must add to 2k, which means

S=2k
S^*=2(k-2)

But taking the differences now yields

S-S^*=2k-2(k-2)=4

and there is only one k for which k-1=4; namely, k=5. However, the sum of the even integers between 1 and 5 is 2+4=6, whereas 2k=10\neq6. So there are no solutions to this over the odd integers.
5 0
3 years ago
A boat travels at a speed of 20 miles per hour in still water. It travels 48 miles upstream, and then returns to the starting po
Zanzabum

Answer:

pepepopo

Step-by-step explanation:

ubsbsvxvxvxvxvxb b

red

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3 0
4 years ago
Convert 689 into presentage
AnnZ [28]

68900%

Step-by-step explanation:

wherever the decimal point is you place the percent sign two places over

5 0
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