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3 years ago

HELP PLSSSSS PLSSSS I NEED HELPPPPPPP ASAPPPPP

Mathematics

1 answer:

Hitman42 [59]3 years ago

5 0

I would say A (7) but I’m not 100% sure.

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What is greater 8 yd or 288 in

natali 33 [55]

To solve this problem we need to know the equivalence between yards and inches, that is:
1 yd = 36 inches

Therefore in 8 yd we have:
8(36) = 288
so in 8 yd we have 288 inches
So 8 yd is equal to 288 in

3 0

3 years ago

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The equation of the line of best fit of a scatter plot is y = 6x − 9. What is the slope of the equation?

nignag [31]

The slope of the equation is 6

8 0

3 years ago

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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

Consider the line y= -7/4x+8

VARVARA [1.3K]

Perpendicular line has slope that multiplies to -1
paralel line has same slope

y=mx+b
slope=-7/4
perpendicular
4/7
(-7,-2)
-2=4/7(-7)+b
-2=-4+b
add 4 both sides
2=b
y=4/7x+2

paralel
y=-7/4x+b
-2=-7/4(-7)+b
-2=49/4+b
minus 49/4 from both sides (-2=-8/4)
-57/4=b
y=-7/4x-14.25

perpendicular
y=4/7x+2
paralel
y=-7/4x-14.25

8 0

3 years ago

Help quick please, look at the photo I attached

Tema [17]

Answer:

A ≈ 5.6

Step-by-step explanation:

The area (A) of the triangle is calculated as

A = $\frac{1}{2}$ × PQ × PR × sin P

= 0.5 × 5.2 × 2.8 × sin50°

≈ 5.6 ( to 1 dec. place )

3 0

3 years ago