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2 years ago

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If a = 2.3 units, b = 4 units, c = 6 units, and d = 8 units, what is the volume of the two prisms?

1 answer:

mihalych1998 [28]2 years ago

5 0

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HELP PLZ AGAIN. Same material. It is still timed. Will give 50 pts and brainly. This is vital to my math grade. Time is on the e

Sliva [168]

Answer:

slope-intercept form: y = x+2

The slope is 1

y-interspet is -2

Step-by-step explanation:

hope it helps

7 0

2 years ago

Solve the equation<br><br> c/7=6 1/7

Softa [21]

c/7 = 6 1/7 ⇒ c = 43 ⇒ 43/7 = 6 1/7

Convert mixed fraction into a simple fraction

6 1/7 ⇒ (6 * 7 + 1)/7 = 43/7

c/7 = 43/7

proportionality:

a/b = c/d where ad = bc

a = c
b = 7
c = 43
d = 7

ad = bc ⇒ c7 = 7*43 ⇒ c7 = 301 ⇒ 7c/7 = 301/7 ⇒ c = 43

5 0

3 years ago

Read 2 more answers

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

Find the value of x so that the figures have the same area.<br><br> 25ft, x-6 ft<br><br> x ft, 15ft

levacccp [35]

X=11ft
25-6=14
So ur answer is actually 3

3 0

3 years ago

I have having a problem with this equation 3(-2-3x) = -9x-4

inn [45]

Answer:

3(-2-3x) = 9x-4

-6-9x = 9x-4

-6+6-9x = 9x-4+6

-9x = 9x+2

-9x-9x = 9x-9x+2

-18x = 2

-18x÷(-18) = 2÷(-18)

x = -(1/9)

Thank you

BY GERALD GREAT

6 0

2 years ago