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borishaifa [10]
2 years ago
11

30 points pls helppp

Mathematics
1 answer:
Phoenix [80]2 years ago
3 0

Answer:

(y + 1) = 3( x +2)

Step-by-step explanation:

from equation,

y - y1 = m( x - x1)

given x1=2, y1= -1

m = 3 (slope)

substitute

y - (-1) = 3( x - 2)

ans: y+1= 3(x+2)

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Three numbers are in the ratio 3:9:10. If 10 is added to the last number, then the three numbers form an arithmetic progression.
MrRa [10]

Answer:

<em> The numbers are 6, 18, and 30 </em>

Step-by-step explanation:

If the three numbers are in the ratio of 3:9:10,

let the numbers be 3x, 9x and 10x.

<em>If 10 is added to the last number to form an arithmetic progression</em>

<em>Then, 3x 9x (10x+10) are the progression</em>

The common difference of an arithmetic progression (d) = T₂ - T₁ = T₃ - T₂

T₂-T₁ = T₃ - T₂ .............. Equation 1

Where T₁ = first term of the progression, T₂ = Second term of the progression, T₃ = third term of the progression

<em>Given: T₁ = 3x, T₂ = 9x, T₃ = 10x +10</em>

<em>Substituting these values into equation 1</em>

<em>9x-3x = (10x+10)-9x</em>

<em>Solving the equation above,</em>

<em>3x = 10+x</em>

<em>3x-x = 10</em>

<em>2x = 10</em>

<em>x = 10/2</em>

<em>x = 2.</em>

<em>Therefore the numbers are 6, 18, and 30 </em>

<em />

7 0
4 years ago
A special liquid is held in a tank described as (x 2 + y 2 ) ≤ z ≤ 1 in a Cartesian coordinate system. Assume that the density o
vivado [14]

Since \rho=\dfrac mV (density = mass/volume), we can get the mass/weight of the liquid by integrating the density \rho(x,y,z) over the interior of the tank. This is done with the integral

\displaystyle\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^1(2-z^2)\,\mathrm dz\,\mathrm dy\,\mathrm dx

which is more readily computed in cylindrical coordinates as

\displaystyle\int_0^{2\pi}\int_0^1\int_{r^2}^1(2-z^2)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{\frac{3\pi}4}

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3 years ago
What proportion of students are willing to report cheating by other students? A project put this question to an SRS of 180 under
krok68 [10]

Answer:

13.89% of students are willing to report cheating by other students.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 180

Number of students who reported cheating, x = 25

We have to find the proportion of the students are willing to report cheating by other students.

Proportion of students can be calculate as

p = \dfrac{x}{n} = \dfrac{25}{180} = 0.1389\\\\\p = 0.1389\times 100\5\\p = 13.89\%

Thus, 13.89% of students are willing to report cheating by other students.

8 0
3 years ago
Find the exact value of cos(sin^-1(-5/13))
son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}

\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

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Answer:

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Step-by-step explanation:

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