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Dmitriy789 [7]
2 years ago
14

Need helpppp pleaseeeee

Mathematics
1 answer:
Temka [501]2 years ago
4 0

Umm...

What is your question??? You have no question in the question box or attachments.

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Element X is a radioactive isotope such that every 25 years, its mass decreases by half. Given that the initial mass of a sample
Julli [10]
50*(.5)^(23/25)=26.43
7 0
2 years ago
I understand some of this but this ones got me stumped i tried pemdas but im still lost
Mandarinka [93]
Just replace x with its given value:

9x^{2}-4x-3~~~~\text{where }x=-2\\ \\ \\ 9\cdot (-2)^{2}-4\cdot (-2)-3\\ \\ =9\cdot 4+8-3\\ \\ =36+8-3\\ \\ =41
6 0
3 years ago
There are 6 minutes of commercials for every 24 minutes of television.How many minutes of commercials are there in 1hour and 36
LUCKY_DIMON [66]

commercial time 6mins per 24 mins

in 1h 36 m (=96mins) there are 4 - 24 minute periods. Therefore there are 4 x6mins of commercial time = 24mins

7 0
3 years ago
BRAINIEST!!! only answer if you know and can give an explanation, will report for non-sense answers
sergiy2304 [10]

Answer:

Below

Step-by-step explanation:

For a given shape to be a rhombus, it should satisfy these conditions:

● The diagonals should intercept each others in the midpoint.

● The diagonals should be perpendicular.

● The sides should have the same length.

We will prove the conditions one by one.

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let's prove that the diagonals are perpendicular:

To do that we will write express them as vectors

The two vectors are EG and DF.

The coordinates of the four points are:

● E(0,2c)

● G (0,0)

● F (a+b, c)

● D (-a-b, c)

Now the coordinates of the vectors:

● EG (0-0,0-2c) => EG(0,-2c)

● DF ( a+b-(-a-b),c-c) => DF (2a+2b,0)

For the diagonals to be perpendicular the scalar product of EG and DF should be null.

● EG.DF = 0*(2a+2b)+(-2c)*0 = 0

So the diagonals are perpendicular.

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let's prove that the diagonals intercept each others at the midpoints.

The diagonals EG and DF should have the same midpoint.

● The midpoint of EG:

We can figure it out without calculations. Since G is located at (0,0) and E at (0,2c) then the distance between E and G is 2c.

Then the midpoint is located at (0,c)

● The midpoint of DF:

We will use the midpoint formula.

The coordinates of the two points are:

● F (a+b,c)

● D(-a-b,c)

Let M be the midpoint of DF

●M( (a+b-a-b,c+c)

● M (0,2c)

So EG and DF have the same midpoint.

■■■■■■■■■■■■■■■■■■■■■■■■■■

There is no need to prove the last condition, since the two above guarante it.

But we can prove it using the pythagorian theorem.

8 0
3 years ago
20 – 10 + 5x = 40 What value of x makes the equation true? *<br> explanition plz
Arlecino [84]

Answer:

6

Step-by-step explanation:

If you solve for x you get your answer. First, 20 - 10 gets you 10. Then you subtract that 10 from both sides of your equation so it now looks like 5x = 30. Last you divide both sides by 5 and get x = 6.

6 0
2 years ago
Read 2 more answers
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