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2 years ago

Find the eighth term of P.G 1,2,4,8

2 answers:

3 0

$a_n=2^{n-1}\\\\a_1=2^{1-1}=2^0=1\\\\a_2=2^{2-1}=2^1=2\\\\a_3=2^{3-1}=2^2=4\\\\a_4=2^{4-1}=2^3=8\\\vdots\\a_{8}=2^{8-1}=2^7=128$

MAVERICK [17]2 years ago

3 0

the answer is 128, the pattern is multiplying the previous number by 2 so example: 2x2=4 4x2=8 etc.

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Find the present balance.

Triss [41]

<h3>Credited balance:-</h3>

$\\ \rm\longmapsto 115.90+115.90+345.85$

$\\ \rm\longmapsto 231.80+345.85$

$\\ \rm\longmapsto 577.65$

<h3>Debited amount:-</h3>

$\\ \rm\longmapsto 25.85+2.00$

$\\ \rm\longmapsto 27.85$

<h3>Checked amount:-</h3>

$\\ \rm\longmapsto 19.45+21.02+95.98$

$\\ \rm\longmapsto 117+19.45$

$\\ \rm\longmapsto 136.45$

<h3>Total change in balance</h3>

$\\ \rm\longmapsto 577.65-(27.85+136.45)$

$\\ \rm\longmapsto 577.65-191.3$

$\\ \rm\longmapsto 386.35$

Now

<h3>Current balance</h3>

$\\ \rm\longmapsto 386.35+271.31$

$\\ \rm\longmapsto\$ 657.66$

8 0

2 years ago

Trig proofs with Pythagorean Identities.

lorasvet [3.4K]

To prove:

$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1$

Solution:

$$LHS = \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}$

Multiply first term by $\frac{1+cos x}{1+cos x}$ and second term by $\frac{1-cos x}{1-cos x}$ .

$$= \frac{1(1+\cos x)}{(1-\cos x)(1+\cos x)}-\frac{\cos x(1-\cos x)}{(1+\cos x)(1-\cos x)}$

Using the identity: $(a-b)(a+b)=(a^2-b^2)$

$$= \frac{1+\cos x}{(1^2-\cos^2 x)}-\frac{\cos x-\cos^2 x}{(1^2-\cos^2 x)}$

Denominators are same, you can subtract the fractions.

$$= \frac{1+\cos x-\cos x+\cos^2 x}{(1^2-\cos^2 x)}$

Using the identity: $1-\cos ^{2}(x)=\sin ^{2}(x)$

$$= \frac{1+\cos^2 x}{\sin^2x}$

Using the identity: $1=\cos ^{2}(x)+\sin ^{2}(x)$

$$=\frac{\cos ^{2}x+\cos ^{2}x+\sin ^{2}x}{\sin ^{2}x}$

$$=\frac{\sin ^{2}x+2 \cos ^{2}x}{\sin ^{2}x}$ ------------ (1)

$RHS=2 \cot ^{2} x+1$

Using the identity: $\cot (x)=\frac{\cos (x)}{\sin (x)}$

$$=1+2\left(\frac{\cos x}{\sin x}\right)^{2}$

$$=1+2\frac{\cos^{2} x}{\sin^{2} x}$

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

Equation (1) = Equation (2)

LHS = RHS

$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1$

Hence proved.

5 0

3 years ago

Simplify the expression given below. x+2/4x^2+5x+1 * 4x+1/x^2-4

GenaCL600 [577]

Answer:

$\frac{1}{(x+1)(x-2)}$

Step-by-step explanation:

$\frac{x+2}{4x^2 + 5x + 1} \ \times \ \frac{4x+1}{x^2-4}\\\\=\frac{x+2}{4x^2 + 4x + x + 1} \ \times \ \frac{4x+1}{x^2-2^2}\\\\=\frac{x+2}{4x(x + 1) + 1( x + 1)} \ \times \ \frac{4x+1}{(x - 2)(x + 2)} \ \ \ \ \ \ \ \ [ \ (a^2 - b^2 = (a-b)(a+b) \ ]\\\\\\=\frac{x+2}{(4x + 1)(x+1)} \ \times \ \frac{4x+1}{(x-2)(x+2)}\\\\=\frac{1}{(x+1)} \ \times \ \frac{1}{(x-2)}\\\\= \frac{1}{(x+1)(x-2)}$

8 0

2 years ago

Find the midpoint (6,-7) and (4,1)

shepuryov [24]

Answer:

(5, -3)

Step-by-step explanation:

let (x1, y1) = (6, -7) and (x2, y2) = (4,1)

(Xmidpoint, Ymidpoint) = ( (x1+x2) / 2, (y1+y2) / 2)

= (5, -3)

5 0

2 years ago

Use a surface integral to find the general formula for the surface area of a cone with height latex: h and base radius latex: a(

BlackZzzverrR [31]

We can parameterize this part of a cone by

$\mathbf s(u,v)=\left\langle u\cos v,u\sin v,\dfrac hau\right\rangle$

with $0\le u\le a$ and $0\le v\le2\pi$ . Then

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=\sqrt{1+\dfrac{h^2}{a^2}}u\,\mathrm du\,\mathrm dv$

The area of this surface (call it $\mathcal S$ ) is then

$\displaystyle\iint_{\mathcal S}\mathrm dS=\sqrt{1+\frac{h^2}{a^2}}\int_{v=0}^{v=2\pi}\int_{u=0}^{u=a}u\,\mathrm du\,\mathrm dv=a^2\sqrt{1+\frac{h^2}{a^2}}\pi=a\sqrt{a^2+h^2}\pi$

6 0

2 years ago