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Deffense [45]
3 years ago
5

The square of a positive number increased by twice the number equals three. Find the number

Mathematics
1 answer:
koban [17]3 years ago
8 0

Answer:

1

Step-by-step explanation:

Let the number be x

x² + 2x = 3

x² + 2x - 3 = 0

Product  = -3

Sum =2

Factors = 3 * (-1)

x² + 3x - x - 3 = 0

x*(x+3) - (x + 3) = 0

(x + 3)(x - 1) = 0

Ignore x + 3 because the number is a positive number

x - 1 = 0

x = 1

The number is 1

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David rolls a fair dice 240 times. <br> How many times would David expect to roll a three?
jolli1 [7]

Answer:

40

Step-by-step explanation:

There are 6 numbers on an dice (1-6) Divide 240 by 6 and the reasonable amount of times to roll a 3 is 40 time

4 0
3 years ago
In the formula C = 2tr, C stands for
maks197457 [2]

Answer:

circumference of a circle

Step-by-step explanation:

Actually, the formula is C = 2πr.  C stands for "circumference."

3 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
Solve the following y=x-5 and y=5x-13<br><br> a) (-4,5) <br> b) (-3,2)<br> c) (2,-3)<br> d) (2,3)
valentinak56 [21]
Answer is C
The x and y hit at (2,-3)
7 0
3 years ago
Find a, b, and c please and thank u
storchak [24]
A=80
B=100
C= 11
Is this the I-ready
8 0
2 years ago
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