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professor190 [17]
3 years ago
11

Y=4x−9 Complete the missing value in the solution to the equation

Mathematics
1 answer:
igomit [66]3 years ago
4 0

Answer:

The possibilities for solutions are pretty much endless, but an example of one would be (1, -5).

Step-by-step explanation:

You can plug in any value for x and then solve for y or plug in any value for y and then solve for x.

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How many pairs of whole numbers have the sum of 99
valentina_108 [34]
50 pairs of whole numbers
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A rectangle has a perimeter of 48 cm and a width of 6 cm. Use the formula P=2ℓ+2w to find the length.
Vinil7 [7]

Answer:the length is equal to 18. 2l+2w= 12+?=48. 48-12=36, 36/2=18

Step-by-step explanation:

5 0
3 years ago
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Explain why a change from 20 to 40 is a 100% increase, but a change from 40 to 20 is a 50% decrease.
Rainbow [258]

Answer:

Because 20 has to be doubled to become 40. While 40 is twice as much as 20 so you just divide 40 by 20 to get 50%.

Explanation:

20 x 20 = 40 = 100%

40 / 20 = 50%

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3 years ago
How many different anagrams can you make for the word MATHEMATICS?
Alex73 [517]

Answer:

4989600

Step-by-step explanation:

The given word is MATHEMATICS

Number of letters in the given word is 11

there are 2 M's, 2 A's and 2 T's

the given word is arranged in 11! ways

three letters are repeating, so we divide by repeating occurrences

number of different arrangements =\frac{11!}{2!2!2!}

= 4989600

7 0
3 years ago
Y''+y'+y=0, y(0)=1, y'(0)=0
mars1129 [50]

Answer:

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form ay''+by'+cy=0.

Given: y''+y'+y=0

Let y=e^{rt} be it's solution.

We get,

\left ( r^2+r+1 \right )e^{rt}=0

Since e^{rt}\neq 0, r^2+r+1=0

{ we know that for equation ax^2+bx+c=0, roots are of form x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} }

We get,

y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}

For two complex roots r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta, the general solution is of form y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )

i.e y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Applying conditions y(0)=1 on e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right ), c_1=1

So, equation becomes y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

On differentiating with respect to t, we get

y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )

Applying condition: y'(0)=0, we get 0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}

Therefore,

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

3 0
3 years ago
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