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3 years ago

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 ...

$\boxed{x=\sqrt{w(w+z)}}$

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3 years ago

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3 years ago

Which expression is equivalent to 4z+2z

Pepsi [2]

Answer:6z

Step-by-step explanation:

4z+2z=6z

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3 years ago

What is the vertex of the function (X) = -x + 71 - 4?

4vir4ik [10]

Answer:

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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:

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

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3 years ago