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GaryK [48]
2 years ago
10

Percentages just answer the sentance

Mathematics
2 answers:
Whitepunk [10]2 years ago
5 0

Answer:

20

Step-by-step explanation:

70%×20

70÷100×20

70÷5

=14

almond37 [142]2 years ago
4 0

Answer:

70% of 20 peaches is 14 peaches

20 is the answer

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at middleton middle school, marianne musf score and average of at least 80 points of 4 tests before she can apply for the schola
Colt1911 [192]
The answer is 83, using the equation of 320-79-81-77=x and X is equal to 83
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3 years ago
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Simplify the expression (2−3i) (5+i) by distributing. Express your answer in a+bi form, Step by Step.
Usimov [2.4K]

Answer:

13-13i

Step-by-step explanation:

2(5+i)-3i(5+i)

10+2i-15i-3i²

10-13i+3

13-13i

7 0
3 years ago
Power Series Differential equation
KatRina [158]
The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for y

\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0

which indeed gives the recurrence you found,

a_{n+3}=-\dfrac{n-3}{n+3}a_n

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that a_2=0, and substituting this into the recurrence, you find that a_2=a_5=a_8=\cdots=a_{3k-1}=0 for all k\ge1.

Next, the linear term tells you that 6a_0+6a_3=0, or a_3=a_0.

Now, if a_0 is the first term in the sequence, then by the recurrence you have

a_3=a_0
a_6=-\dfrac{3-3}{3+3}a_3=0
a_9=-\dfrac{6-3}{6+3}a_6=0

and so on, such that a_{3k}=0 for all k\ge2.

Finally, the quadratic term gives 6a_1-12a_4=0, or a_4=\dfrac12a_1. Then by the recurrence,

a_4=\dfrac12a_1
a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1
a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1
a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1

and so on, such that

a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}

for all k\ge2.

Now, the solution was proposed to be

y=\displaystyle\sum_{n\ge0}a_nx^n

so the general solution would be

y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots
y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)
y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)
4 0
3 years ago
Calculate the area of a parallelogram with a base of 3 feet and a height of 1.2 feet
Fantom [35]

Answer: 3.6

Step-by-step explanation: 3x1.2= 3.6

3 0
3 years ago
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I need help real quick please!
iogann1982 [59]

f(x) - n - move the graph n units down

f(x) + n - move the graph n units up

f(x - n) - move the graph n units right

f(x + n) - move the graph n units left

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f(x)=x^2\\\\g(x)=f(x)-3=x^2-3

<h3>Answer: g(x) = x² - 3</h3>
6 0
3 years ago
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