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3 years ago

5

All the edges of the object in the diagram are equal in length. The object is cut by a vertical plane containing A and B and bis

ecting two of the horizontal edges. What is the shape of the cross-section resulting from the cut?

2 answers:

SashulF [63]3 years ago

7 0

Answer:

square

Step-by-step explanation:

PLATO kids unite!!!

inn [45]3 years ago

3 0

Its a rhombus I believe ..<span />

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Khan sucks 3 for 30 points<br> again

tester [92]

Answer:

D, John did not make a mistake!

Step-by-step explanation:

7 0

3 years ago

Sladkaya [172]

D. x > 5
Combine like terms. 3x - x = 2x and 3 - 7 = -4. The equation is now 2x - 4 > 6. Now add 4 to both sides. 2x > 10. Now divide by 2 to get 'x' alone, leaving you with x > 5

4 0

3 years ago

Read 2 more answers

In the figure, TS = 8, SR = 4, and QR = 2, what is the length of PQ?

Sauron [17]

Answer:

d. 22

Step-by-step explanation:

(PQ + QR) × QR = (TS + SR) × SR => two secants intersecting theorem

PQ = ?

QR = 2

TS = 8

SR = 4

Plug in the values

(PQ + 2) × 2 = (8 + 4) × 4

2PQ + 4 = 12 × 4

2PQ + 4 = 48

2PQ + 4 - 4 = 48 - 4

2PQ = 44

2PQ/2 = 44/2

PQ = 22

5 0

3 years ago

SOLVE. integration of (1+v^2) /(1-v^3)

s2008m [1.1K]

$\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv$

$1-v^3=(1-v)(1+v+v^2)$
$\implies\dfrac{1+v^2}{1-v^3}=\dfrac a{1-v}+\dfrac{b_0+b_1v}{1+v+v^2}$
$\implies\dfrac{1+v^2}{1-v^3}=\dfrac{a(1+v+v^2)+(b_0+b_1v)(1-v)}{1-v^3}$
$\implies 1+v^2=(a+b_0)+(a-b_0+b_1)v+(a-b_1)v^2$
$\implies\begin{cases}a+b_0=1\\a-b_0+b_1=0\\a-b_1=1\end{cases}\implies a=\dfrac23,b_0=\dfrac13,b_1=-\dfrac13$

So,

$\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv=\dfrac23\int\frac{\mathrm dv}{1-v}+\dfrac13\int\frac{1-v}{1+v+v^2}\,\mathrm dv$

The first integral is easy. For the second, since the derivative of the denominator is $(1+v+v^2)=1+2v$ , we can add and subtract $3v$ to get

$\dfrac{1-v}{1+v+v^2}=\dfrac{1+2v-3v}{1+v+v^2}=\dfrac{1+2v}{1+v+v^2}-\dfrac{3v}{1+v+v^2}$

and for the first term employ a substitution. For the remaining term, we can complete the square in the denominator, then use a trigonometric substitution:

$\displaystyle\int\frac{1+2v}{1+v+v^2}\,\mathrm dv=\int\frac{\mathrm dt}t$

where $t=1+v+v^2\implies\mathrm dt=(1+2v)\,\mathrm dv$ , and

$\displaystyle\int\frac{3v}{1+v+v^2}\,\mathrm dv=3\int\frac v{\left(v+\frac12\right)^2+\frac34}\,\mathrm dv$

Then taking $v+\dfrac12=\dfrac{\sqrt3}2\tan s\implies \mathrm dv=\dfrac{\sqrt3}2\sec^2s\,\mathrm ds$ gives

$\displaystyle3\int\frac v{\left(v+\frac12\right)^2+\frac34}\,\mathrm dv=3\int\frac{\frac{\sqrt3}2\tan s-\frac12}{\left(\frac{\sqrt3}2\tan s\right)^2+\frac34}\left(\frac{\sqrt3}2\sec^2s\right)\,\mathrm ds$
$=\displaystyle\sqrt3\int(\sqrt3\tan s-1)\,\mathrm ds$
$=\displaystyle3\int\tan s\,\mathrm ds-\sqrt3\int\mathrm ds$

Now we're ready to wrap up.

$\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv=\dfrac23\int\frac{\mathrm dv}{1-v}+\dfrac13\int\frac{1-v}{1+v+v^2}\,\mathrm dv$
$=\displaystyle-\frac23\ln|1-v|+\frac13\left(\int\frac{1+2v}{1+v+v^2}\,\mathrm dv-\int\frac{3v}{1+v+v^2}\,\mathrm dv\right)$
$=\displaystyle-\frac23\ln|1-v|+\frac13\int\frac{\mathrm dt}t-\frac13\left(3\int\tan s\,\mathrm ds-\sqrt3\int\mathrm ds\right)$
$=\displaystyle-\frac23\ln|1-v|+\frac13\ln|t|-\int\tan s\,\mathrm ds+\frac1{\sqrt3}\int\mathrm ds$
$=\displaystyle-\frac23\ln|1-v|+\frac13\ln|1+v+v^2|+\ln|\cos s|+\frac s{\sqrt3}+C$
$=\displaystyle-\frac23\ln|1-v|+\frac13\ln|1+v+v^2|+\ln\left|\frac{\sqrt3}{2\sqrt{1+v+v^2}}\right|+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C$

This can be simplified a bit using some properties of logarithms to obtain

$=\displaystyle-\frac23\ln|1-v|+\frac13\ln(1+v+v^2)+\left(\ln\frac{\sqrt3}2-\frac12\ln(1+v+v^2)\right)+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C$
$=\displaystyle-\frac23\ln|1-v|-\frac16\ln(1+v+v^2)+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C$

6 0

3 years ago

The mean mark of a group of 10 boys is 58. If the mean of 7 of them is 61, what is the mean of the remaining 3 boys

Nesterboy [21]

Answer:

The mean of the remaining 3 boys is 51.

Step-by-step explanation:

The mean mark of the entire group ( $p_{10}$ ), dimensionless, is:

$p_{10} = \frac{1}{10}\cdot \Sigma_{i = 1}^{10} x_{i}$ (Eq. 1)

( $p_{10} = 58$ )

$\frac{1}{10}\cdot \Sigma_{i=1}^{10} x_{i} = 58$ (Eq. 1b)

Where $x_{i}$ is the mark of the i-th boy, dimensionless.

In addition, we know the following mean marks from statement:

$p_{7} = \frac{1}{7} \cdot \Sigma_{i = 1}^{7} x_{i}$ (Eq. 2)

( $p_{7} = 61$ )

$\frac{1}{7}\cdot \Sigma_{i=1}^{7} x_{i} = 61$ (Eq. 2b)

$p_{3} = \frac{1}{3}\cdot \Sigma_{i=8}^{10}x_{i}$ (Eq. 3)

Where:

$p_{7}$ - Mean mark of the first 7 boys, dimensionless.

$p_{3}$ - Mean mark of the remaining 3 boys, dimensionless.

By applying sum properties in (Eq. 1b) and using (Eq. 2b) and (Eq. 3), we obtain the mean of the remaining 3 boys:

$\frac{1}{10}\cdot [\Sigma_{i = 1}^{7}x_{i}+\Sigma_{i=8}^{10}x_{i}] = 58$

$\frac{1}{10}\cdot [7\cdot (61)+3\cdot p_{3}] = 58$

$7\cdot (61) + 3\cdot p_{3} = 580$

$3\cdot p_{3} = 153$

$p_{3} = \frac{153}{3}$

$p_{3} = 51$

The mean of the remaining 3 boys is 51.

7 0

3 years ago