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klasskru [66]
3 years ago
15

Find the perimeter of a rectangle with length 6 units and width 4 units. wa

Mathematics
1 answer:
agasfer [191]3 years ago
5 0

Answer:

The perimeter of the given rectangle  = 20 units.

Step-by-step explanation:

Length of the rectangle =  6 units

The width of the rectangle = 4 units

Now Perimeter of the Rectangle  = 2 ( LENGTH + WIDTH)

⇒Here, perimeter  = 2 ( 6 units + 4 units ) =  2 x ( 10 units)

                                 = 20 units

Hence, the perimeter of the given rectangle  = 20 units.

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Don't know how to express the value of X as any of those answer choices
Alekssandra [29.7K]

9514 1404 393

Answer:

  x = √(w(w+z))

Step-by-step explanation:

The idea with a right-triangle figure of this sort is that all of the triangles are similar. Here, x is the short side of ∆ABC, and the hypotenuse of ∆BDC. This suggests you want to write a similarity statement involving the short side and hypotenuse.

  BC/AC = DC/BC . . . . . short side/hypotenuse

  BC² = AC·DC . . . . . . cross multiply

  x² = (w+z)w . . . . . . substitute letter values

Taking the square root and rearranging to the form of the applicable answer choice, this is ...

  \boxed{x=\sqrt{w(w+z)}}

3 0
3 years ago
How do I find angle a?
vfiekz [6]
The slope of the line is 2/3, so the angle is arctan(2/3) ≈ 33.69°.
8 0
3 years ago
Which expression is equivalent to 4z+2z
Pepsi [2]

Answer:6z

Step-by-step explanation:

4z+2z=6z

4 0
3 years ago
What is the vertex of the function (X) = -x + 71 - 4?
4vir4ik [10]

Answer:

Step-by-step explanation:

f(x) = -x + 71 - 4 is a straight line. It does not have a vertex.

7 0
3 years ago
Choose the best coordinate system to find the volume of the portion of the solid sphere rho <_4 that lies between the cones φ
MrRissso [65]

Answer:

So,  the volume is:

\boxed{V=\frac{128\sqrt{2}\pi}{3}}

Step-by-step explanation:

We get the limits of integration:

R=\left\lbrace(\rho, \varphi, \theta):\, 0\leq \rho \leq  4,\, \frac{\pi}{4}\leq \varphi\leq \frac{3\pi}{4},\, 0\leq \theta \leq 2\pi\right\rbrace

We use the spherical coordinates and  we calculate a triple integral:

V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_0^4  \rho^2 \sin \varphi \, d\rho\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \left[\frac{\rho^3}{3}\right]_0^4\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \cdot \frac{64}{3} \, d\varphi\, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} [-\cos \varphi]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}  \, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\

we get:

V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\V=\frac{64\sqrt{2}}{3}\cdot[\theta]_0^{2\pi}\\\\V=\frac{128\sqrt{2}\pi}{3}

So,  the volume is:

\boxed{V=\frac{128\sqrt{2}\pi}{3}}

4 0
3 years ago
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