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3 years ago

Hello guys how y’all doing

Mathematics

1 answer:

Basile [38]3 years ago

6 0

Answer: I’m good.

Step-by-step explanation: Need Help?

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10.

True [87]

Answer:

Good question. That's a hard one but I think it's C. {(-4,-6), (-3,-2), (1,-2), (1,0)}

Step-by-step explanation:

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3 years ago

A function is negative over the intervals

rewona [7]

Answer:

-3 and 3

Step-by-step explanation:

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3 years ago

Given OSALE, solve for x. 3 3x + 4 5x-6 S 3 A

murzikaleks [220]

Answer:

x=5

Step-by-step explanation:

The sides have to be equal length

3x+4 = 5x-6

Subtract 3x from each side

3x+4-3x = 5x-6-3x

4 = 2x-6

Add 6 to each side

4+6 = 2x-6+6

10 = 2x

Divide by 2

10/2 =2x/2

5 =x

5 0

3 years ago

Read 2 more answers

the equation is -2x² = 4-3 (x + 1) and the question is justify that it is a 2nd degree equation with the unknown x complete.

Serggg [28]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the equation

-2x² = 4-3 (x + 1)

-2x² = 4-3x-3

-2x² = -3x -7

0 = 2x² -3x -7

We know that the degree of the equation is the highest power of x variable in the given equation.

In the equation 0 = 2x² -3x -7 the highest power of x variable in the given equation is 2.

Thus, the degree of the equation is 2.

Also in the equation 0 = 2x² -3x -7, the unknown variable is 'x'.

Let us determine the value 'x'

2x² -3x -7 = 0

Add 7 to both sides

$2x^2-3x-7+7=0+7$

$2x^2-3x=7$

Divide both sides by 2

$\frac{2x^2-3x}{2}=\frac{7}{2}$

$x^2-\frac{3x}{2}=\frac{7}{2}$

Add (-3/4)² to both sides

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{7}{2}+\left(-\frac{3}{4}\right)^2$

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{65}{16}$

$\left(x-\frac{3}{4}\right)^2=\frac{65}{16}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{3}{4}=\sqrt{\frac{65}{16}}$

$x-\frac{3}{4}=\frac{\sqrt{65}}{\sqrt{16}}$

$x-\frac{3}{4}=\frac{\sqrt{65}}{4}$

Add 3/4 to both sides

$x-\frac{3}{4}+\frac{3}{4}=\frac{\sqrt{65}}{4}+\frac{3}{4}$

$x=\frac{\sqrt{65}+3}{4}$

similarly solving

$x-\frac{3}{4}=-\sqrt{\frac{65}{16}}$

$x=\frac{-\sqrt{65}+3}{4}$

So the solution of the equation will have the values of x such as:

$x=\frac{\sqrt{65}+3}{4},\:x=\frac{-\sqrt{65}+3}{4}$

6 0

3 years ago

Write down the general zeroth order linear ordinary differential equation. Write down the general solution.

nikitadnepr [17]

The zeroth derivative of a function $y(x)$ is simply the function itself, so the zeroth order linear ODE takes the general form

$y(x)=f(x)$

whose solution is $f(x)$ .

7 0

3 years ago