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3 years ago

Please help!! i missed my hw deadline

Mathematics

1 answer:

Paha777 [63]3 years ago

4 0

Bruh i ain't got time for that Goodluck tho

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What number multiples to -1 and adds to -50

koban [17]

Z: is the number

[Z*(-1)]+(-50)
(-Z)+(-50)
-Z-50

<span>I hope it helps you :)</span>

5 0

2 years ago

The cake store is having a 25% off sale on all of its cakes.

Shalnov [3]

Answer:

$6.75

Step-by-step explanation:

First you find 25% of $9----> 2.25 and then you subtract that from 9----> 9-2.25=<u>6.75</u>

4 0

2 years ago

Read 2 more answers

A statement with the word "per" is always a unit rate. true/false

11111nata11111 [884]

Answer:

True

Step-by-step explanation:

The definition of per is for each. Per is used with units to express a rate.

3 0

2 years ago

Use the standard normal distribution to find P(-1.23 Sz<2.12).

matrenka [14]

Answer:

(d) 0.8736

Step-by-step explanation:

Unless you have a table of the normal distribution available, this is a calculator problem. Your calculator, or any spreadsheet, can tell you the probability is ...

P(-1.23 ≤ z ≤ 2.12) ≈ 0.8736

5 0

1 year 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

2 years ago