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Elza [17]
3 years ago
5

Prove angle BCD is congruent to angle FED

Mathematics
1 answer:
mash [69]3 years ago
7 0
With no given information, we're not sure that it actually is.
We don't even know for sure whether the two angles are
even on the same planet.
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A miner has 3 kilograms of gold dust. She needs to share it evenly with four partners. How much gold should each of the five peo
djverab [1.8K]
600 g

Explanation:
Since 1 kilogram =1000 grams
Therefore 3 grams= 3000 grams
Now divide it with 5 partners
3000 divided by 5=600 grams

Each person will get 600 grams of gold dust
6 0
3 years ago
Barb is making a bead necklace. She strings 1 white bead,then 3 blue beads,then 1 white bead,and so on. Write the numbers for th
nataly862011 [7]

The numbers for the first eight beads that are white are; 1, 5, 9, 13, 17, 21, 25, 29

<h3>Arithmetic Sequence</h3>

We are told that she strings in this order;

  • 1 white
  • 3 blue
  • 1 white bead

Now, after the first white bead, the next white bead is the 5th bead in total.

This means the difference in interval between white beads is; 5 - 1 = 4 beads

Thus, the numbers for the first eight beads that are white are;

1, (1 + 4), (1 + 4 + 4), (1 + 4 + 4 + 4).....

Thus, we have;

1, 5, 9, 13, 17, 21, 25, 29

Read more on arithmetic sequence at; brainly.com/question/7849474

8 0
2 years ago
Read 2 more answers
3x+y=14<br> Rewrite equation in slope-intercept form(solve y)
attashe74 [19]
Y = -3x + 14
Subtracted 3x from both sides
6 0
3 years ago
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If a = √3-√11 and b = 1 /a, then find a² - b²​
Serga [27]

If b=\frac1a, then by rationalizing the denominator we can rewrite

b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8

Now,

a^2 - b^2 = (a-b) (a+b)

and

a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8

a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8

\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}

5 0
2 years ago
PLEASE HELP ME!!!!! I need Help Please With the 2nd question please any help me
galina1969 [7]
Hello,
Please, see the attached file.
Thanks.

3 0
3 years ago
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