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natita [175]
2 years ago
13

1) A lake near the Arctic Circle is covered by a 2-meter-thick sheet of ice during the cold winter

Mathematics
1 answer:
Delvig [45]2 years ago
6 0
The answer to your question is c
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1/7 divided by 4 equals
gulaghasi [49]
1/ 28 is the answer.
= 1/7 ÷4
= 1/7 ÷ 4/1
=1/7 X 1/4 
Multiply top by top, bottom by bottom
=1/28 

If you do not know how to do it:
First make both sides into a fraction ( Whole numbers always have a 1 as a denominator).
Then you reciprocate (Flip) the second fraction.
Finally you multiply top by top, bottom by bottom
You get the answer!<span />
6 0
3 years ago
PLEASE SOMEONE HELP ASAP
Elan Coil [88]
The line is 13 units long
5 0
3 years ago
Read 2 more answers
The perimeter of a triangle is 39 feet. One side of the triangle is one foot longer than the second side. The third side is two
Ivenika [448]

Answer:

Step-by-step explanation:

39 feet for one side

40 feet for another

42 feet for the last side

I got the answer by adding 1 foot to the first side which comes out to be 40 feet, I got the last answer by adding 2 feet to 40 which is 42.

3 0
3 years ago
What is reminder if you devide 786 to 6?​
Ivenika [448]

Answer:

There is no reminder as the answer will be an integer 131.

Thank you and please rate me as brainliest as it will help me to level up

5 0
3 years ago
Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec
julia-pushkina [17]

Answer:

(2,6,6) \not \in \text{Span}(u,v)

(-9,-2,5)\in \text{Span}(u,v)

Step-by-step explanation:

Let b=(b_1,b_2,b_3) \in \mathbb{R}^3. We have that b\in \text{Span}\{u,v\} if and only if we can find scalars \alpha,\beta \in \mathbb{R} such that \alpha u + \beta v = b. This can be translated to the following equations:

1. -\alpha + 3 \beta = b_1

2.2\alpha+4 \beta = b_2

3. 3 \alpha +2 \beta = b_3

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for \alpha,\beta and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

-\alpha + 3 \beta = 2

2\alpha+4 \beta = 6

whose unique solutions are \alpha =1 = \beta, but note that for this values, the third equation doesn't hold (3+2 = 5 \neq 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

-\alpha + 3 \beta = -9

2\alpha+4 \beta = -2

whose unique solutions are \alpha=3, \beta=-2. Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.

4 0
2 years ago
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