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nikklg [1K]
2 years ago
12

Brainliest if answer good

Mathematics
1 answer:
nasty-shy [4]2 years ago
4 0
Your not showing the multiple choice and I can’t see it good
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Diedra has $20 to spend. Can she buy shorts that cost $18.50?
Anit [1.1K]

Answer:

yes she can

Step-by-step explanation:

yes she can because $20 is great than $18.50

8 0
2 years ago
Read 2 more answers
Would you rather receive $10,000 a day for 31 days or a penny that doubles in value every day for 31 days?
bagirrra123 [75]

This shows how much you would make per day for both choices.

Total earned in 31 days for $10,000 per day: $310,000

Explanation

$10,000 every day in 31 days.

Total earned in 31 days for penny: 21,474,836.47

Explanation

Day 1: $0.01

Day 2: $0.02

Day 3: $0.04

Day 4: $0.08

Day 5: $0.16

Day 6: $0.32

Day 7: $0.64

Day 8: $1.28

Day 9: $2.56

Day 10: $5.12

Day 11: $10.24

Day 12: $20.48

Day 13: $40.96

Day 14: $81.92

Day 15: $163.84

Day 16: $327.68

Day 17: $655.36

Day 18: $1,310.72

Day 19: $2,621.44

Day 20: $5,242.88

Day 21: $10,485.76

Day 22: $20,971.52

Day 23: $41,943.04

Day 24: $83,886.08

Day 25: $167,772.16

Day 26: $335,544.32

Day 27: $671,088.64

Day 28: $1,342,177.28

Day 29: $2,684,354.56

Day 30: $5,368,709.12

Day 31: $10,737,418.24

7 0
2 years ago
<img src="https://tex.z-dn.net/?f=z%5E%7B7%7D%3D128i" id="TexFormula1" title="z^{7}=128i" alt="z^{7}=128i" align="absmiddle" cla
abruzzese [7]

If <em>z</em> ⁷ = 128<em>i</em>, then there are 7 complex numbers <em>z</em> that satisfy this equation.

z^7 = 128i = 2^7i = 2^7e^{i\frac\pi2}

\implies z=\sqrt[7]{2^7} e^{i\frac17\left(\frac\pi2+2n\pi\right)}

(where <em>n</em> = 0, 1, 2, …, 6)

\implies z = 2 e^{i\left(\frac\pi{14}+\frac{2n\pi}7\right)}

\displaystyle\implies z = 2 \left(\cos\left(\frac\pi{14}+\frac{2n\pi}7\right)+i\sin\left(\frac\pi{14}+\frac{2n\pi}7\right)\right)

3 0
3 years ago
Which of the operations +, -, x, ÷ are inverses of each other?
Verizon [17]
+ & -
X & divide
Hope this helps!
6 0
3 years ago
Read 2 more answers
What are the possible numbers of positive, negative, and complex zeros of f(x) = −x6 − x5 − x4 − 4x3 − 12x2 + 12?
ad-work [718]
f(x) = -x^6 -x^5 - x^4 - 4x^3 - 12x^2 + 12

There is only one change of sign, so there is only one possible positive root.

f(-x) = -(-x)^6 -(-x)^5 - (-x)^4 - 4(-x)^3 - 12(-x)^2 + 12\\&#10;f(-x) = -x^6 +x^5 - x^4 + 4x^3 - 12x^2 + 12

There are five changes of signs, so there are 5,3 or 1 possible negative roots.

The number of complex roots can be equal to 4,2 or 0 (degree of a polynomial - possible positive roots - possible negative roots)

8 0
3 years ago
Read 2 more answers
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