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3 years ago

How many units is 1 from -6 on a number line

1 answer:

8 0

Find this by subtracting the two terms and then finding the absolute value of the difference.

1-(-6)= 7
|7|= 7

Final answer: 7

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.......Help Please.........

Sveta_85 [38]

Answer:

57% to the nearest whole number

Step-by-step explanation:

That is (80/139.3) * 100

= 0.5743 * 100

= 57.43%.

8 0

3 years ago

I need help because I don’t know what they are asking

Mashutka [201]

Answer:

16x -2

Step-by-step explanation:

You know how to compute the perimeter of a rectangle of length L and width W:

P = 2(L+W)

Here, you're asked to use the given algebraic expressions for length and width and simplify the result of putting those in the perimeter formula.

P = 2((5x-2) +(3x+1))

P = 2(8x -1)

P = 16x -2 . . . the perimeter of the rectangle

6 0

3 years ago

A(n)=−5+6(n−1)<br> What is the 12th term in the sequence

barxatty [35]

Answer:

n/2[2a+(n-1)d]

12/2[2(-5)+(12-1)6]

66-10=56×6=336

336 is the 12th term

4 0

2 years ago

Two times a number minus six another number is six. Two times the first number added to three times the other number is twenty f

expeople1 [14]

Answer:

First Number: 2

Step-by-step explanation:

You need to work backwards, as an algebraic equation.

24 ÷ 3 = 8

8 ÷ 2 = 4

4 × 2 = 8

8 - 6 = 2

The first number is 2.

- Educationist

4 0

2 years ago

The general solution of 2 y ln(x)y' = (y^2 + 4)/x is

Sav [38]

Replace $y'$ with $\dfrac{\mathrm dy}{\mathrm dx}$ to see that this ODE is separable:

$2y\ln x\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2+4}x\implies\dfrac{2y}{y^2+4}\,\mathrm dy=\dfrac{\mathrm dx}{x\ln x}$

Integrate both sides; on the left, set $u=y^2+4$ so that $\mathrm du=2y\,\mathrm dy$ ; on the right, set $v=\ln x$ so that $\mathrm dv=\dfrac{\mathrm dx}x$ . Then

$\displaystyle\int\frac{2y}{y^2+4}\,\mathrm dy=\int\dfrac{\mathrm dx}{x\ln x}\iff\int\frac{\mathrm du}u=\int\dfrac{\mathrm dv}v$

$\implies\ln|u|=\ln|v|+C$

$\implies\ln(y^2+4)=\ln|\ln x|+C$

$\implies y^2+4=e^{\ln|\ln x|+C}$

$\implies y^2=C|\ln x|-4$

$\implies y=\pm\sqrt{C|\ln x|-4}$

4 0

3 years ago