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3 years ago

12

Find theProve that sin theta tan theta= (1 + sec theta)(1 - cos theta)

2 answers:

just olya [345]3 years ago

8 0

Sin tita tan tita = LHS

larisa86 [58]3 years ago

5 0

Answer:

Step-by-step explanation:

RHS = (1 +Sec Ф)(1 - Cos Ф)

= 1*1 - 1*Cos Ф + Sec Ф *1 - Sec Ф *Cos Ф

= 1 - Cos Ф + Sec Ф - 1 {Sec Ф = $\frac{1}{Cos \ theta}$ )

=Sec Ф - Cos Ф

= $\frac{1}{Cos \ theta}$ - Cos Ф

$= \frac{1}{Cos \ theta}-\frac{Cos \ theta*Cos \ theta}{Cos \ theta}\\\\= \frac{1-Cos^{2} \ theta}{Cos \ theta}\\\\= \frac{Sin^{2} \ theta}{Cos \ theta}\\\\=\frac{Sin \ theta*Sin \ theta}{Cos \ theta}\\\\= Sin \ theta*\frac{Sin \ theta}{Cos \ theta}$

= Sin Ф* tan Ф = LHS

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MArishka [77]

Answer: 60 degrees

Step-by-step explanation: triangles are come to 180 if you add all angles correctly. therefore if you do 180-90-30, you'll get 60.

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2 years ago

Can someone please find the volume of this shape and show work for how you did it??

castortr0y [4]

Answer:

$221.4m^{3}$

Step-by-step explanation:

To work out the volume of this shape you will first need to work out the area of the triangle. To work out the area of the triangle you would multiply the base of 12 by the height of 4.1, which gives you 49.2. You would then need to divide 49.2 by 2, which gives you 24.6. This is because the formula for the area of a triangle is the same as that of a square, however a triangle is half of a square with the same length and width. To work out the volume you would multiply the area of the triangle, which is 24.6 by the height of 9, which gives you 221.4 $m^{3}$

1) Multiply 12 by 4.1.

$12*4.1=49.2$

2) Divide 49.2 by 2.

$49.2/2=24.6$

3) Multiply 24.6 by 9.

$24.6*9=221.4m^{3}$

6 0

2 years ago

<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0

2 years ago

Math question down below

Alja [10]

Answer:

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3 years ago

timurjin [86]

Answer:

s> 144 2/3

Step-by-step explanation:

I took the test in k12 and i got it right so yea !

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1 year ago