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zhuklara [117]
3 years ago
9

Find the value of r in the triangle shown below. ​

Mathematics
1 answer:
jok3333 [9.3K]3 years ago
5 0
The answer is A. Hope that helps
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Please help ASAP :{ jhjkhybblvklmmm
babunello [35]

Answer:

<h2>C</h2>

Step-by-step explanation:

V=0.5*b*l*h

V=0.5*10*28*12

V=10*14*12

V=10*168

V=1680 m^3 or C

8 0
3 years ago
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As the value of the multiple coefficient of determination increases,
Leona [35]

Answer:

a.the goodness of fit for the estimated multiple regression equation increases.

Step-by-step explanation:

As the value of the multiple coefficient of determination increases,

a. the goodness of fit for the estimated multiple regression equation increases.

As we know that the coefficient of determination measures the variability of response variable with the help of regressor. As we know that if the value of the coefficient of determination increases strength of fit also increases.  

7 0
3 years ago
Melinda bought 6 bowls for $13.20. what was the unit rate in dollars
never [62]
<u>6
</u>13.20
Divide by 6
<u>1
</u>2.2
6 0
3 years ago
PLEASE HELP I NEED THE ANSWER ASAP
Andreas93 [3]

Answer:

x>3, the last option with the open circle

Step-by-step explanation:

7 0
3 years ago
Find a particular solution to y" - y + y = 2 sin(3x)
leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

     =e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}

Multiplying both sides of equation (1) by integrating factor and integrating we get

\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I

I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx

8 0
3 years ago
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