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goldfiish [28.3K]

2 years ago

6

I have a question if Jimmy has 3 apples and is friend ate 12 apples off a tree and jimmy eats one and then his friend (Sihera) p

icks 12 new apples and gives two to jimmy nd eats 4 of them how many apples does jimmy and Sihera have?
I made this question :D

1 answer:

Roman55 [17]2 years ago

4 0

13?

Step-by-step explanation:

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PLEASE HELP ME!!!!!!! I think the answer is exponential but I am not 100% sure!!!!!!!

babunello [35]

It is exponentially each time you are dividing by 4/3

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3 years ago

Given: B bisects EA, FE is parallel to CA, DC is parallel

SIZIF [17.4K]

Answer:

Choice 3: AAS

Step-by-step explanation:

We can prove that by AAS that means we need two congruent angles and one congruent side.

The first angle will be the vertical pair <FBG and <DBC.

The second angle will be the alternate interior pair <G and <D.

The one side will be $\overline{GB}$ and $\overline{BD}$ .

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2 years ago

-5y y=-2 -3x 6y=-12 substitution

IrinaK [193]

The answer would be x=4 and y=-2.

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3 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

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

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1 year ago

Is this relation a function justify your answer

Karo-lina-s [1.5K]

No because there is absolutely no correlation

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2 years ago

Read 2 more answers