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Svetach [21]
3 years ago
5

Help please asap I’m stuck

Mathematics
1 answer:
alexandr402 [8]3 years ago
5 0

Answer:

A rhombus has diagonals which bisect each other making both diagonal be cut in 2 equal parts

Therefore the halved parts of the diagonal have equal measurements

Meaning 2x+4 = 7x-1

Then solve for x from there

Please give me brainliest answer :)

Step-by-step explanation:

You might be interested in
The number of bacteria in a colony growing exponentially. At 4 pm yesterday the number of bacteria is 400.And at 6 pm yesterday
Doss [256]

Answer:

10800

Step-by-step explanation:

The number of bacteria in a colony growing exponentially. At 4 pm yesterday the number of bacteria is 400.And at 6 pm yesterday it was 3600. How many bacteria in the colony at 7 pm?

Step-by-step explanation:

B = B₀ Kⁿ

At 4 pm yesterday the number of bacteria is 400

B₀ = 400

n = number of hrs

at 6 pm yesterday it was 3600 => n =  6- 4 = 2 hrs

B₂ = 3600

=> 3600 = 400 * k²

=> k² =  9

=> k = 3

At 7 Pm Yesterday

n = 7-4 = 3

B₃ = 400 * 3³

=> B₃ = 27 * 400

=> B₃ = 10800

bacteria in the colony at 7 pm = 10800

or B₂ = 3600 & n = 7-6 = 1

Then B₃ = 3600 * 3¹ = 10800

And boom here

5 0
3 years ago
Read 2 more answers
one bouquet of flowers has 3 roses for every 5 carnations. other bouquet has 4 carnations for every 5 daisies. if both bouquets
Leno4ka [110]

Answer:

therefore the second bouquet will have more flowers , as it will have 25 daisies

Step-by-step explanation:

3 roses for every 5 carnations

5 daisies for every 4 carnations

we can solve this using cross multiplication

3 : 5

x : 20

5x = 60

x = 60/5

<em>x = 12</em>

meaning that for a bouquet that has 20 carnations there will be 12 roses

5 : 4

x : 20

4x = 100

<em>x = 25</em>

meaning that for a bouquet that has 20 carnations there will be 25 daisies.

<u>therefore the second bouquet will have more flowers , as it will have 25 daisies</u>

6 0
3 years ago
Marcia claims that the GCF for (2x2 + 4xy + 8xy4) is 8x2y4.
miskamm [114]
She is not correct.
The GCF for 2x^2 + 4xy + 8xy^4
is 2x
The first term does not have a 'y' in it so you cannot have a 'y' in the GCF.
The second and third terms both only have 1 'x', so you can only take out 1 'x', not 2.
The highest constant that all three terms have in common is a 2.
8 0
4 years ago
Please help. I think I know but I'd like to have it confirmed.
V125BC [204]
Part 1
40 \times 10.5 = 402
Part 2
40 \times 10.5 + 5 \times 1.5 \times 10.5 \\ = 402 + 78.75 \\ = 480.75
4 0
3 years ago
Helppppppppppppppppppppppppp
Klio2033 [76]
  1. 6+t=6+t
  2. y+11
  3. 12-4=8
  4. 7/t
  5. 25-19=6
  6. 2s+s
  7. 11-r
  8. 2s+6x
  9. x/17-(12+2)
  10. 3+(x-10)
  11. (7+2)/2
  12. x-13

Hope this helps and have a BLESSED DAY! I also put some charts with my answer to help you better understand it!

-Cutiepatutie :-)

8 0
3 years ago
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