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Nitella [24]
3 years ago
14

3.

Mathematics
1 answer:
Ber [7]3 years ago
6 0
A. 7x

B. 8x-1

C. 6x+12

D. 3x +4

E. 13x-7
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Solve the inequalities by graphing. Select the correct graph.<br>y&gt; X<br>y&lt;-X​
Veseljchak [2.6K]

Answer:

5.3

Step-by-step explanation:

7 0
3 years ago
What's the probability of the compound event for. A bag contains 4 red, 8 blue, 6 yellow and 2 green marbles. Maria selects a ma
Nataly [62]

Given:

Number of red marbles = 4

Number of blue marbles = 8

Number of yellow marbles = 6

Number of green marbles = 2

To find:

The probability of getting a blue marble then a red marble, i.e., P(blue, red).

Solution:

Using the given information,

The total number of marbles = 4+8+6+2

                                                = 20

Probability of getting a blue marble in first draw is

P(Blue)=\dfrac{\text{Number of blue marbles}}{\text{Total number of marbles}}

P(Blue)=\dfrac{8}{20}

P(Blue)=\dfrac{2}{5}

Maria selects a marble, puts it back and then selects a second marble. It means the total number of marbles remains the same.

P(red)=\dfrac{\text{Number of red marbles}}{\text{Total number of marbles}}

P(red)=\dfrac{4}{20}

P(red)=\dfrac{1}{5}

Now, the probability of getting a blue marble then a red marble is

P(blue,red)=\dfrac{2}{5}\times \dfrac{1}{5}

P(blue,red)=\dfrac{2}{25}

Therefore, the required probability P(blue, red) is \dfrac{2}{25}.

5 0
3 years ago
Logarithmic to Exponential
jeka57 [31]

A = B \log_x C\\\\\text{Given that,}~~ A = 11, ~B = 4~ C = 190\\\\~~~~~~~~11 = 4 \log_x (190)\\\\\implies \log_x (190) = \dfrac{11}4\\\\\implies x^{\tfrac{11}4} = 190\\\\\implies x = 190^{\tfrac 4{11}}\\\\ \implies x \approx 6.7397

4 0
2 years ago
How do you simplify 2m²×2m³ ?​
Rainbow [258]

Answer:

2m² (2m)

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin
saw5 [17]
Treat \mathcal C as the boundary of the region \mathcal S, where \mathcal S is the part of the surface z=2xy bounded by \mathcal C. We write

\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r

with \mathbf f=(y+7\sin x,z^2+9\cos y,x^3).

By Stoke's theorem, the line integral is equivalent to the surface integral over \mathcal S of the curl of \mathbf f. We have


\nabla\times\mathbf f=(-2z,-3x^2,-1)

so the line integral is equivalent to

\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S
=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv


where \mathbf s(u,v) is a vector-valued function that parameterizes \mathcal S. In this case, we can take

\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)

with 0\le u\le1 and 0\le v\le2\pi. Then

\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv

and the integral becomes

\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv
=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi<span />
4 0
3 years ago
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