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3 years ago

Organize for whether it is equivalent to 9/8, 8/9, or neither.

Mathematics

2 answers:

Aneli [31]3 years ago

5 0

18/16, 1/8x9, - 9/8 group 16/18, 8/9, 1/18x16 - 8/9 group 1/8x1/9- neither

MariettaO [177]3 years ago

5 0

The ones that go in 9/8 group are 18/16 and 1/8 × 9.
The ones that go into 8/9 group are 1/18 × 16, 8/9 and 16/18.

Unfortunately, 1/8 × 1/9 does not fit in any of these groups.So it comes in neither..!

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I got stuck, I don’t even know where to start I think the answer is A but idk. Help pls

kolbaska11 [484]

Answer:

yeah, it's A your correct!

hope it helps;)

4 0

3 years ago

There are 120 men and 380 women in the disco live show. What is the ratio of women to men in the show?

Paladinen [302]

Answer:

6:19

Step-by-step explanation:

There are 120 men and 380 women.

The ratio of women to men in the show is 120:380.

Simplify 120:380.

⇒ 6:19

6 0

4 years ago

Read 2 more answers

PLEASE HELP!! SHOW WORK! <3<br> Find the area of this figure. Dimensions are in meters.

densk [106]

Answer:

2√46 + 25π/4 ≈ 33.2 . . . . m²

Step-by-step explanation:

The altitude of the triangle is given by the Pythagorean theorem. The right triangle of interest is the one that has 5√2 as its hypotenuse, and a leg of half the base shown. Then the other leg, the altitude of the triangle, is ...

h = √((5√2)² - 2²) = √46

Then the area of the triangle shown is ...

A = (1/2)bh = (1/2)(4)(√46)

A = 2√46

The area of the semicircle is given by the formula ...

A = (1/2)πr²

Filling in the radius shown, the area is computed as ...

A = (1/2)π(5√2/2)² = 25π/4

So, the total area of the figure is ...

total area = triangle area + semicircle area

= 2√46 + 25π/4 . . . square meters

≈ 33.2 . . . square meters

3 0

3 years ago

Someone help me!!!please

gregori [183]

9514 1404 393

Answer:

115 square units

Step-by-step explanation:

The area of a parallelogram is the product of base length and height. The height is the perpendicular distance between the bases.

A = bh

A = (23)(5) = 115 . . . . square units

8 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