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3 years ago

What could be the intersection of the plane PRS and plane QRS

1 answer:

6 0

The plane PRS passes through the points P, R and S. So it contains the line RS. Also the plane QRS passes through the points Q, R and S. So it contains the line RS as well. Since both the planes contain the line RS, the line RS must be the intersection of plane PRS and QRS

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Prove that an = 4^n + 2(-1)^nis the solution to

olga nikolaevna [1]

Answer:

See proof below

Step-by-step explanation:

We have to verify that if we substitute $a_n=4^n+2(-1)^n$ in the equation $a_n=3a_{n-1}+4a_{n-2}$ the equality is true.

Let's substitute first in the right hand side:

$3a_{n-1}+4a_{n-2}=3(4^{n-1}+2(-1)^{n-1})+4(4^{n-2}+2(-1)^{n-2})$

Now we use the distributive laws. Also, note that $(-1)^{n-1}=\frac{1}{-1}(-1)^n=(-1)(-1)^{n}$ (this also works when the power is n-2).

$=3(4^{n-1})+6(-1)^{n-1}+4(4^{n-2})+8(-1)^{n-2}$

$=3(4^{n-1})+(-1)(6)(-1)^{n}+4^{n-1}+(-1)^2(8)(-1)^{n}$

$=4(4^{n-1})-6(-1)^{n}+8(-1)^{n}=4^n+2(-1)^n=a_n$

then the sequence solves the recurrence relation.

4 0

3 years ago

Jayne says that 0.45 is equal to 4/10. Is she correct? Explain.

yKpoI14uk [10]

No, 4/10 is equal to 0.40 and not 0.45 since 4/10 goes in tens.

4 0

3 years ago

Read 2 more answers

HELP, Prove: (cos(x))/(sin(x)tan(x))=cot^2x

Mice21 [21]

Answer:

Steps shown below

Step-by-step explanation:

We will simplify this using definitions and identities. Let's start.

$\frac{cos(x)}{sin(x)tan(x)}$

Using $tanx=\frac{sinx}{cosx}$ , we have:

$\frac{cos(x)}{sin(x)\frac{sin(x)}{cos(x)}}\\=\frac{cos^2(x)}{sin^2(x)}$

Using $cot(x)=\frac{cos(x)}{sin(x)}$ , we have:

$\frac{cos^2(x)}{sin^2(x)}\\=cot^2(x)$

Hence, proved.

7 0

3 years ago

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Pythagoream theorem <br> a=18 b=26 c=?

Arisa [49]

A^2 + b^2 = c^2
18^2 + 26^2 = c^2
324 + 676 = c^2
1000 = c^2
sqrt 1000 = c
31.62 = c

5 0

3 years ago

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.10.A cone with a height of 15 yards has a volume of 457.17 yd3 . Find the diameter of the cone

Neko [114]

$\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=457.17\\ h = 15 \end{cases}\implies 457.17=\cfrac{\pi r^2(15)}{3}\implies 457.17=5\pi r^2 \\\\\\ \cfrac{457.17}{5\pi }=r^2\implies \sqrt{\cfrac{457.17}{5\pi }}=r\implies 5.39 \approx r~\hfill \boxed{\stackrel{diameter = 2r}{2(5.39) = 10.78}}$

7 0

4 years ago