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Rina8888 [55]
2 years ago
11

Which word phrase represents the numerical expression below? * 4+(27-10)

Mathematics
1 answer:
lorasvet [3.4K]2 years ago
6 0

Answer:

4 plus the difference of 27 and 10

Step-by-step explanation:

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I need the incomplete fraction answer for:<br> y=<br> z=<br> b=<br> Thx
sergejj [24]

Answer:

Y = 15sqrt(3)/4

Z = 15sqrt(3)/2

b = 45/4

Step-by-step explanation:

Sin(60) =Z/15

sqrt(3)/2 = Z/15

Z = 15sqrt(3)/2

Sin(30) = Y/Z

½ = Y/Z

Y = ½(15sqrt(3))/2 = 15sqrt(3)/4

Cos(30) = b/Z

sqrt(3)/2 = b/Z

b = 15sqrt(3)/2 × sqrt(3)/2

b = 45/4

sqrt is square root/radical

8 0
2 years ago
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Factor p^2-q^2-2p+2q
Grace [21]
Not sure of the answer but i know what factors
p ^2 - 2p

p (p-2)

-q^2 +2q

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so maybe (p-q) (p-2) (q-2)
6 0
2 years ago
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Order from least to greatest 61% , 0.605, 3/5, 59%
krek1111 [17]
So you'd have to convert all of the values to one form, and I find it easiest to do it in decimals. 61% would be 0.61, 0.605 stays the same, 3/5 would be 0.6, and 59% would be 0.59. Now you can order them:
0.59, 0.6, 0.605, 0.61
You have to convert them back to their original form, however, so your answer would be
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6 0
3 years ago
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Which of the following is equivalent to <br><br> 5t−2r=25
lidiya [134]

Answer:

Solving for t : t=5+2r/5

Solving for r : r=-25/2+5t/2

Step-by-step explanation: Move all of your terms that don't contain r to the right side and solve. Move all terms that don't contain t to the right side and solve.

Hope this helps you out! ☺ I just solved it and I don't have any choices to go off of and so I just solved it. Sorry if it's wrong. ☺

3 0
3 years ago
Let E = {(x, y) e R2|xy &gt; 0}. Determine whether E is a subspace of R2 . X3
gayaneshka [121]

Answer:

E is not a subspace of \mathbb{R}^2

Step-by-step explanation:

E is not a subspace of  \mathbb{R}^2

In order to see this, we must find two points (a,b), (c,d) in  E such that (a,b) + (c,d) is not in E.

Consider

(a,b) = (1,1)

(c,d) = (-1,-1)

It is easy to see that both (a,b) and (c,d) are in E since 1*1>0 and (1-)*(-1)>0.  

But (a,b) + (c,d) = (1-1, 1-1) = (0,0)

and (0,0) is not in E.

By the way, it can be proved that in any vector space all sub spaces must have the vector zero.

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2 years ago
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