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s2008m [1.1K]
3 years ago
11

If a = mg - kv²/ m find , correct to the nearest whole number the value of V when a= 2.8 ,m= 12 ,g = 9.8 and k= 8/3

Mathematics
2 answers:
xxMikexx [17]3 years ago
6 0

Answer:

Step-by-step explanation:

Harlamova29_29 [7]3 years ago
5 0

Answer:

v = 22.59

Step-by-step explanation:

all you need is contained in this sheet

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Please help me with geometry! &lt;3<br> In rhombus ABCD, m&lt;=53*, What is M (*=degrees)
julia-pushkina [17]
In a rhombus opposite angles are equal and all 4 add up to 360
assuming one angle is 53, the opposite angle will also be 53 and then adjacent angles will be (360 - (53 x 2))/2 = 254/2 = 127*(deg)
I think the answer you're looking for is 127
3 0
3 years ago
Please help quickly !
GalinKa [24]

Answer:

1. multiplication property of equality

2. division  property of equality

Step-by-step explanation:

1. multplied by -3/2

2. divided by -2/3

8 0
3 years ago
Wich relation is a function of x
faust18 [17]

Answer:

the first one

Step-by-step explanation:

In the other 2 i can see the x repeats itself meaning it isn't a function.

7 0
3 years ago
Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one sam
avanturin [10]
Let "a" and "s" represent the costs of advance and same-day tickets, respectively. Your problem statement gives you two relations.
.. a + s = 35 . . . . . the combined cost of one of each is 35
.. 15a +40s = 900 . . total paid for this combination of tickets was 900

There are many ways to solve these equations. You've probably been introduced to "substitution" and "elimination" (or "addition"). Using substitution for "a", we have
.. a = 35 -s
.. 15(35 -s) +40s = 900 . . substitute for "a"
.. 25s +525 = 900 . . . . . . . simplify
.. 25s = 375 . . . . . . . . . . . .subtract 525
.. s = 15 . . . . . . . . . . . . . . .divide by 25
Then
.. a = 35 -15 = 20

The price of an advance ticket was 20.
The price of a same-day ticket was 15.
6 0
4 years ago
For each of the solutions of the equations find two consecutive integers between which the solution is located.
adoni [48]

Using squares of integers numbers, it is found that the solution of the equation is located between the integers x = 1 and x = 2.

The equation given is:

x^2 = 3

The solution of the equation given is:

x = \sqrt{3}

The squares of the integers numbers until the square root of 3 are:

1^2 = 1

2^2 = 4

Since 1 < 3 < 4, the square root of 3, which is the solution to the equation, is located between the integers x = 1 and x = 2.

A similar problem is given at brainly.com/question/3729492

4 0
3 years ago
Read 2 more answers
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