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3 years ago

Plz help me I am struggling

Mathematics

2 answers:

irakobra [83]3 years ago

6 0

I think it is not true because all sides are triangle except the one on the bottom is a square

sasho [114]3 years ago

6 0

It is possible because the square part of the pyramid is flat and has pointed edges,
The pyramid has one point so...

Flat then pointed

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Fill the blank with <,>or = -25.5___-20.5

tangare [24]

Step-by-step explanation:

<h3>Hope it helps you !!!</h3>

7 0

2 years ago

Verify if 1/2x-2 and g(x) = 2x + 4 are inverses

Triss [41]

They are inverses if f( (g(x)) = x

f((g(x)) = (1/2) (2x + 4) - 2 = x + 2 - 2 = x

so this proves that g(x) is the inverse of f(x)

the other inverse g((f(x)) should also be = x

5 0

2 years ago

What is the solution to 3x+30+x=10+2x+5x+2?

solmaris [256]

Answer:

answer is x = 6

„„„„„„„„»»»»»»»»»»

I hope it helps you

:-))

4 0

3 years ago

Read 2 more answers

0.611 convert Decimal to percentage

diamong [38]

Answer:

0.611 = 61.1%

Step-by-step explanation:

'Percent (%)' means 'out of one hundred':

p% = p 'out of one hundred',

p% = p/100 = p ÷ 100.

Note:

100/100 = 100 ÷ 100 = 100% = 1

Multiply a number by the fraction 100/100,

... and its value doesn't change.

Calculate the percent value:

0.611 =

0.611 × 100/100 =

(0.611 × 100)/100 =

61.1/100 =

61.1%;

In other words:

1) Multiply that number by 100.

2) Add the percent sign % to it.

7 0

2 years ago

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$

3 0

1 year ago