All categories

3 years ago

20 POINTS!!

Mathematics

2 answers:

kirill115 [55]3 years ago

4 0

multiply -2/25 x 5 hours

-2/25 x 5/1 = -10/25 = -2/5

Rzqust [24]3 years ago

3 0

Answer:

It is - 2/5

Step-by-step explanation:

You might be interested in

A circle with circumference 130 has an arc with a 72°<br> angle. What is the length of the arc?

deff fn [24]

Answer:

arc = 26 units

Step-by-step explanation:

130 = 360°

(72° arc/360°)(72) = 26

4 0

2 years ago

The graph of g(x) is the graph of f(x)=x+6 reflected across the x-axis.Which equation describes g?

Mandarinka [93]

Reflected across x axis means y turns to -y, meaning, multiply the whole function by -1

f(x) times -1=-1 times (x+6)=-x-6

A is answer

3 0

3 years ago

One of the factors of 1331x to the 3rd power-8y to the 3rd power.

artcher [175]

Remember
a^3-b^3=(a-b)(a^2+ab+b^2)

(11x)³-(2y)³=(11x-2y)(121x²+22xy-4y²)

6 0

3 years ago

Perform the indicated operation. X^3+43-49x-7x^2 divides by x^2-49

BaLLatris [955]

Answer:

The quotient is: x-7

and remainder is : -300

Step-by-step explanation:

We need to divide $X^3+43-49x-7x^2$ by $x^2-49$

First arrange the term $X^3+43-49x-7x^2$ in terms of ascending order of x.

Arranging we get:

$x^3 - 7x^2 -49x + 43$ \ $x^2-49$

The division steps are shown in figure attached.

The quotient is: x-7

and remainder is : -300

4 0

2 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

2 years ago