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3 years ago

15

For which fraction pairs would it be useful to use the benchmarks 0, One-half, and 1 to compare? Check all that apply.

1 answer:

koban [17]3 years ago

8 0

Answer:

,

Step-by-step explanation:

thank my answer

pls mark brilinias

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Please help with both questions

tresset_1 [31]

Value of n is greater than (-2) and less than or equal to 3.
n = {-1,0,1,2 3}

3x +5 > 16
Subtract 5 from both sides
3x > 16 - 5
3x > 11
x > 11/3
x> 3.67

X= 4 is the smallest value

4 0

2 years ago

Read 2 more answers

Joseph's math teacher challenges him to estimate the height of a pine tree next to the

xxTIMURxx [149]

Answer:

im not good at math

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3 years ago

Fill in the blank in the sentence below with the correct term. The _________ of the numbers -4 and 4 is 4

Ray Of Light [21]

Answer:

Absolute value

Step-by-step explanation:

Absolute value is how many numbers it takes to get back to zero therefore it can never be negative, I think the term absolute value fits the blank perfectly

5 0

3 years ago

The total pay for 12 hours of work at a base rate of (p) per hour plus a temporary raise of $2.50 per hour. How would I write th

Naddika [18.5K]

12P+2.50 HOPE THIS HELPS

4 0

3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago