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3 years ago

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

7 0

3 years ago

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

Read 2 more answers

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%

4 0

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