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Irina-Kira [14]
2 years ago
14

Use each integer only once. You must have each row, column, and diagonal adding up to the magic sum.

Mathematics
2 answers:
Grace [21]2 years ago
8 0
For question number 3
-7, 7, 6, -4,
4, -2, -1, 1
0, 2, 3, -3,
5, -5, -6, 8

For question number 4:
-4, 1, -10, 3,
-9, 2, -3, 0,
5, -8, -1, -6,
-2, -5, 4, -7
mixas84 [53]2 years ago
3 0
Ano bayan parang tanga nman ang hirap
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What is 2 plus 2 if you want to know
Murrr4er [49]
2 plus 2 the answer would be 20
8 0
3 years ago
Read 2 more answers
(4m-n-3). 12n13<br> 8m 20
JulsSmile [24]

Answer:

24n^7/m^16

Step-by-step explanation:

Just simplify and rewrite what ever you get simplify again 2 times and now make the calculation as you see every big number is a whole number and is a pair.

5 0
3 years ago
10) Which two ratios form a proportion? A) 1 : 2 and 4 : 2 B) 1 : 2 and 6 : 3 C) 2 : 1 and 2 : 4 D) 2 : 1 and 6 : 3
Likurg_2 [28]

Answer:

2/1and6/3 because if you do cross multilication they both equal 6

Step-by-step explanation:

8 0
3 years ago
Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1
Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

x'=\frac{dx}{dt}

a)  x’=t–sin(t),  x(0)=1

dx=(t-sint)dt

Apply integral both sides:

\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k

where k is a constant due to integration. With x(0)=1, substitute:

1=0+cos0+k\\\\1=1+k\\k=0

Finally:

x=\frac{t^2}{2} +cos(t)

b) x’+2x=4; x(0)=5

dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\

Completing the integral:

-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}

Solving the operator:

-\frac{1}{2}ln(4-2x)=t+k

Using algebra, it becomes explicit:

x=2+ke^{-2t}

With x(0)=5, substitute:

5=2+ke^{-2(0)}=2+k(1)\\\\k=3

Finally:

x=2+3e^{-2t}

c) x’’+4x=0; x(0)=0; x’(0)=1

Let x=e^{mt} be the solution for the equation, then:

x'=me^{mt}\\x''=m^{2}e^{mt}

Substituting these equations in <em>c)</em>

m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}

Finally:

x=\frac{1}{2} sin(2t)

7 0
3 years ago
Linear Equations with explanation plz
MissTica

Answer:

0, -2

Step-by-step explanation:

.................:)

3 0
3 years ago
Read 2 more answers
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