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8_murik_8 [283]
3 years ago
5

What is 2 to the power of 5?

Mathematics
2 answers:
IgorC [24]3 years ago
8 0

Answer:

the answer is 32

Step-by-step explanation:


Stels [109]3 years ago
6 0

Answer:

32

Step-by-step explanation:

2x2x2x2x2=32

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Match each point with one of the following rational number:
castortr0y [4]
A = -0.7
B = -9/16 (0.5623)
C = -0.4
D = -5/16 (0.3125)

The last option is correct 
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Me Henderson has 2 bouncy-ball vending machines. He buys one bag of the 27-millimeter balls and one bag of the 40- millimeter ba
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20 bouncy balls in each machine
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Please help me! I think this 20 character thing is unnecessary when you have an attachment
PIT_PIT [208]

Answer:

B

Step-by-step explanation:

Also the answer must be under 9 so that rules out option D

Because We know CE = 2 and AD is visibly larger than CE, 6We can rule out option A

And that brings us down to B and C

CB is equal to 6, And I'm no genius, but I'm pretty sure AD is DEFINATLY not equal to 6.

so the answer is B

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3 years ago
Figure EFGH on the grid below represents a trapezoidal plate at its starting position on a rotating machine platform:
Zigmanuir [339]

Answer:

the correct answer is D: they form concentric circles.

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3 years ago
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Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
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