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forsale [732]
3 years ago
5

Please help me with this question I will give brainliest.

Mathematics
1 answer:
Ainat [17]3 years ago
6 0

Answer:

All you need to remember is the rules

Step-by-step explanation:

Let us remember

a to the m power x a to the nth power is = a to the m+n power. (add the exponents)

And

a to the m power ÷ a to the nth power is = a to the m-n power. (subtract the exponents) So

14 to the -4 power x 14 to the 7 power= 14 to the -4+7 which is equal to 14 to the 3rd power

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Which of the following points lies on the graph of the function y = 8x?
nignag [31]
Y = 8^x
For (0, 1): 1 ≠ 8^0 = 1
For (8, 1): 1 ≠ 8^8 = 16,777,216
For (2, 64): 64 = 8^2
For (0, 8): 8 ≠ 8^0 = 1

Therefore, option 3, (2, 64) is the correct answer.
8 0
3 years ago
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Z-49+851=100+21z /////////////////////////////////////////////////
WITCHER [35]

Answer:

z=35.1

Step-by-step explanation:

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Kate borrowed $400 from her friend. She will repay this amount plus 3 percent simple interest after one year. What is the total
tekilochka [14]
The answer is $412.

Let's first calculate simple interest. Simple interest (I) can be expressed as:
I = P * r * t
P - principal
r - rate
t - time period

It is given:
I = ?
P = $400
r = 3% = 0.03
t = 1 year

Therefore:
I = P * r * t = 400 * 0.03 * 1 = 12

The total amount Kate will repay is the principal amount (P) plus 3% simple interest (I):
P + I = 400 + 12 = $412
6 0
3 years ago
How good are you setting goals? 3 sentences for honors
Lunna [17]

Answer:

Be more productive

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This could be a goal

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3 years ago
(1) Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.
Evgesh-ka [11]

Answer:

(1)

Multiplying by 3 both sides of the equality you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

3u  is in the Span of the vectors \{v_1,v_2,v_3\}.

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

Step-by-step explanation:

(1)

As the hint indicates, you know that

u = c_1 v_1 + c_2v_2+c_3v_3

Then, if you multiply both sides of the equality by 3, you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

And that's it. 3u  is in the Span of the vectors \{v_1,v_2,v_3\}

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

4 0
3 years ago
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