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3 years ago

Find two consecutive odd integers whose sum is 116.

1 answer:

7 0

x+(x+2)=116. When simplified 2x+2=116. Now solve 2x=116-2. 2x= 114. Now divide to get x alone by dividing both sides by 2. X= 114÷2= 57. So now 116-57= 59 So, your two number 57 and 59. If you add them will be 116

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Obtain the general solution to the equation. (x^2+10) + xy = 4x=0 The general solution is y(x) = ignoring lost solutions, if any

alukav5142 [94]

Answer:

$y(x)=4+\frac{C}{\sqrt{x^2+10}}$

Step-by-step explanation:

We are given that a differential equation

$(x^2+10)y'+xy-4x=0$

We have to find the general solution of given differential equation

$y'+\frac{x}{x^2+10}y-\frac{4x}{x^2+10}=0$

$y'+\frac{x}{x^2+10}y=4\frac{x}{x^2+10}$

Compare with

$y'+P(x) y=Q(x)$

We get

$P(x)=\frac{x}{x^2+10}$

$Q(x)=\frac{4x}{x^2+10}$

I.F= $e^{\int\frac{x}{x^2+10} dx}=e^{\frac{1}{2}ln(x^2+10)}$

$e^{ln\sqrt(x^2+10)}=\sqrt{x^2+10}$

$y\cdot \sqrt{x^2+10}=\int \frac{4x}{x^2+10}\times \sqrt{x^2+10} dx+C$

$y\cdot \sqrt{x^2+10}=\int \frac{4x}{\sqrt{x^2+10}}+C$

$y\cdot \sqrt{x^2+10}=4\sqrt{x^2+10}+C$

$y(x)=4+\frac{C}{\sqrt{x^2+10}}$

6 0

3 years ago

SOLVE what is J<br> 3j + 4 = 10<br> J =

tigry1 [53]

Answer: The answer is 2

Step-by-step explanation:

3j + 4 = 10 J = 2

3x2=6

6+4= 10

6 0

1 year ago

Read 2 more answers

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Anettt [7]

Answer:

Step-by-step explanation:

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irina [24]

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3 years ago

Vlad [161]

The single transformation from shape A to B is: dilate A by 1/2, followed by a 180 degrees rotation

<h3>How to determine the transformation?</h3>

From the figure, we have:

Shape A is twice the size of shape B
Both shapes can be rotated by 180 degrees in either direction to map onto the other, after dilation.

This means that:

The scale of dilation from shape A is:

k = 1/2

Hence, the single transformation from shape A to B is: dilate A by 1/2, followed by a 180 degrees rotation