All categories

3 years ago

Cuanto ganaran si compran el numero del premio mayor

Mathematics

2 answers:

elixir [45]3 years ago

7 0

Answer:

is this spanish? or the real question

Step-by-step explanation:

Ainat [17]3 years ago

4 0

Answer: ill just get my free points

Step-by-step explanation:

You might be interested in

Can I get help on this please

Misha Larkins [42]

Answer:

5. You can round question a up to 100 + 55 (then do the counting)

5. You can round question b up to 245 - 100 (then do the counting)

You can round question c up to 100 x 5 (then do the counting)

You can round question d up to 100 x 20 (then subtract accordingly)

Please rank Brainliest if this helps!

4 0

3 years ago

If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$

6 0

3 years ago

Please help me<br>simplify<br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B32%20-%20%20%5Csqrt%7B50%7D%20%7D%20" id="TexFormul

Mnenie [13.5K]

Answer:

$\sqrt{32-5\sqrt{2} }$

Step-by-step explanation: I hope it helps :)

$\sqrt{32-\sqrt{50} } \\\\simplify \: \sqrt{50} \:to \:5\sqrt{2} \\\\= \sqrt{32-5\sqrt{2} } \\\\Decimal Form: 4.992888$

5 0

3 years ago

What kind of exponent does this graph have? Help pls ?

otez555 [7]

It’s 42 but Carrier the 2

4 0

3 years ago

Find the area of the trapezoid to the nearest tenth.

erica [24]

Answer:

2.2 metres squared

Step-by-step explanation:

We need to find the area of this trapezoid.

The area of a trapezoid is denoted by:

$A=\frac{(b_1+b_2)h}{2}$ , where $b_1$ and $b_2$ are the parallel bases and h is the height

Here, we already know the lengths of the two bases; they are 0.9 metres and 2.3 metres. However, we need to find the length of the height.

Notice that one of the angles is marked 45 degrees. Let's draw a perpendicular line from top endpoint of the segment labelled 0.9 to the side labelled 2.3. We now have a 45-45-90 triangle with hypotenuse 2.0 metres. As one of such a triangle's properties, we can divide 2.0 by √2 to get the length of both legs:

2.0 ÷ √2 = √2 ≈ 1.414 ≈ 1.4

Thus, the height is h = 1.4 metres. Now plug all these values we know into the equation to find the area:

$A=\frac{(b_1+b_2)h}{2}$

$A=\frac{(0.9+2.3)*1.4}{2}=2.2$

The answer is thus 2.2 metres squared.

<em>~ an aesthetics lover</em>

8 0

3 years ago