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mote1985 [20]
3 years ago
11

- 3√45-√ 125 + √200-√150

Mathematics
2 answers:
andrezito [222]3 years ago
8 0

Answer:

=-80-5√5

this is the answer

hope this helps

Murljashka [212]3 years ago
4 0
-29.4102647752 the answer I think
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2.03+7.89 estimate Please Help Asap
Oduvanchick [21]
The correct answer is 9.92
6 0
3 years ago
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A costume designer bought 1.4 kg of sequince. She used 250 grams to make a crown and 400 grams to make a cape. How many grams of
Salsk061 [2.6K]
Find the total amount in grams.

1 kg = 1000 g

1.4 * 1000 = 1400

So she bought a total of 1400 grams.

Add up the grams she used:

400 + 250 = 650

Subtract it from the total:

1400 - 650 = 750

So 750 grams of sequence were left.
5 0
3 years ago
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What's the algebraic expression for the sum of v and 3
dangina [55]
The sum means addition
v + 3
6 0
3 years ago
Find the measure of x.<br> X<br> 55°<br> 80<br> X<br> = [ ?<br> ]<br> Round to the nearest tenth.
NARA [144]
<h3>Answer:  139.5</h3>

Work Shown:

cos(angle) = adjacent/hypotenuse

cos(55) = 80/x

x*cos(55) = 80

x = 80/cos(55)

x = 139.4757 approximately

x = 139.5

3 0
3 years ago
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How many 'words' can be made from the name ESTABROK with no restrictions
Bumek [7]

The number of ways in which the name 'ESTABROK' can be made with no restrictions is 40, 320 ways.

<h3>How to determine the number of ways</h3>

Given the word:

ESTABROK

Then n = 8

p = 6

The formula for permutation without restrictions

P = n! ( n - p + 1)!

P = 8! ( 8 - 6 + 1) !

P = 8! (8 - 7)!

P = 8! (1)!

P = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 × 1

P = 40, 320 ways

Thus, the number of ways in which the name 'ESTABROK' can be made with no restrictions is 40, 320 ways.

Learn more about permutation here:

brainly.com/question/4658834

#SPJ1

6 0
1 year ago
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