Find the value of X.

What is the volume of a square pyramids

frozen [14]

Answer:

$\large\boxed{V=75\ in^3}$

Step-by-step explanation:

The formula ogf a volume of a pyramid:

$V=\dfrac{1}{3}BH$

B - area of a base

H - height

We have a base length a = 5 in and a height H = 9 in.

In the base we have a square. The formula of an area of a square is:

$B=a^2$

Substitute:

$B=5^2=25\ in^2$

Calculate the volume:

$V=\dfrac{1}{3}(25)(9)=(25)(3)=75\ in^3$

7 0

3 years ago

x2 + y2 − 4x + 12y − 20 = 0 (x − 6)2 + (y − 4)2 = 56 x2 + y2 + 6x − 8y − 10 = 0 (x − 2)2 + (y + 6)2 = 60 3x2 + 3y2 + 12x + 18y −

snow_tiger [21]

For this case, what we must do is fill squares in all the expressions until we find the correct result.
We have then:

x2 + y2 − 4x + 12y − 20 = 0 x2 + y2 − 4x + 12y = 20
x2 − 4x + y2 + 12y = 20
x2 − 4x + (12/2)^2 + y2 + 12y + (-4/2)^2 = 20 + (12/2)^2 + (-4/2)^2
x2 − 4x + (6)^2 + y2 + 12y + (-2)^2 = 20 + (6)^2 + (-2)^2
x2 − 4x + 36 + y2 + 12y + 4 = 20 + 36 + 4
(x − 2)2 + (y + 6)2 = 60

3x2 + 3y2 + 12x + 18y − 15 = 0
x2 + y2 + 4x + 6y − 5 = 0
x2 + y2 + 4x + 6y = 5
x2 + 4x + (4/2)^2 + y2 + 6y + (6/2)^2 = 5 + (4/2)^2 + (6/2)^2
x2 + 4x + (2)^2 + y2 + 6y + (3)^2 = 5 + (2)^2 + (3)^2
x2 + 4x + 4 + y2 + 6y + 9 = 5 + 4 + 9
(x + 2)2 + (y + 3)2 = 18

2x2 + 2y2 − 24x − 16y − 8 = 0
x2 + y2 − 12x − 8y − 4 = 0
x2 + y2 − 12x − 8y = 4
x2 − 12x + (-12/2)^2 + y2 − 8y + (-8/2)^2 = 4 + (-12/2)^2 + (-8/2)^2
x2 − 12x + (-6)^2 + y2 − 8y + (-4)^2 = 4 + (-6)^2 + (-4)^2
x2 − 12x + 36 + y2 − 8y + 16 = 4 + 36 + 16
(x − 6)2 + (y − 4)2 = 56

x2 + y2 + 2x − 12y − 9 = 0
x2 + y2 + 2x - 12y = 9
x2 + 2x + y2 - 12y = 9
x2 + 2x + (2/2)^2 + y2 - 12y + (-12/2)^2 = 9 + (2/2)^2 + (-12/2)^2
x2 + 2x + (1)^2 + y2 - 12y + (-6)^2 = 9 + (1)^2 + (-6)^2
x2 + 2x + 1 + y2 - 12y + 36 = 9 + 1 + 36
(x + 1)2 + (y − 6)2 = 46

7 0

3 years ago

A company rents bicycles to customers. The company charges an initial fee

photoshop1234 [79]

The hourly rate is $3 a hour with a $2 initial fee.

5 0

3 years ago

Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).

I am Lyosha [343]

Take the homogeneous part and find the roots to the characteristic equation:

$y''+4y=0\implies r^2+4=0\implies r=\pm2i$

This means the characteristic solution is $y_c=C_1\cos2x+C_2\sin2x$ .

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form $y_p=ax\cos2x+bx\sin2x$ . Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With $y_1=\cos2x$ and $y_2=\sin2x$ , you're looking for a particular solution of the form $y_p=u_1y_1+u_2y_2$ . The functions $u_i$ satisfy

$u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$

where $W(y_1,y_2)$ is the Wronskian determinant of the two characteristic solutions.

$W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2$

So you have

$u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx$
$u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x$

$u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx$
$u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x$

So you end up with a solution

$u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x$

but since $\cos2x$ is already accounted for in the characteristic solution, the particular solution is then

$y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x$

so that the general solution is

$y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x$

7 0

3 years ago

Simplify for x. (10-5 2/3)×(-2)+x=8 2/3

polet [3.4K]

Answer:

16/37 I just did this... answer mines pls

Step-by-step explanation:

3 0

3 years ago

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