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3 years ago

Which of the following answers is the simplified form of 3^2 x 3^-6?

Mathematics

2 answers:

4vir4ik [10]3 years ago

7 0

Answer:

where are the answer choices?

Step-by-step explanation:

dem82 [27]3 years ago

4 0

Answer:

It is 3^-4

Step-by-step explanation:

-6+2=-4

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What basic trigonometric identity would you use to verify that sin^2x +cos^2x/cos x = sec x

gogolik [260]

Answer:

The basic identity used is $\bold{\sin ^{2} x+\cos ^{2} x=1}$ .

Solution:

In this problem some of the basic trigonometric identities are used to prove the given expression.

Let’s first take the LHS:

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

Step one:

The sum of squares of Sine and Cosine is 1 which is:

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

On substituting the above identity in the given expression, we get,

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

Step two:

The reciprocal of cosine is secant which is:

$\cos x=\frac{1}{\sec x}$

On substituting the above identity in equation (1), we get,

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

Thus, RHS is obtained.

Using the identity $\sin ^{2} x+\cos ^{2} x=1$ , the given expression is verified.

6 0

2 years ago

In the figure below, BCA ~ STR. Find cos C, sin C, and tan C. Round your answers to the nearest hundredth.

arsen [322]

Answer:

$\sin C\approx0.85\\\\\cos C \approx0.52\\\\\tan C\approx1.63$

Step-by-step explanation:

According to the trigonometric ratios in aright triangle :

$\sin x =\dfrac{\text{Side opposite to x}}{\text{Hypotenuse}}\\\\\cos x =\dfrac{\text{Side adjacent to x}}{\text{Hypotenuse}}\\\\\tan x=\dfrac{\sin x}{\cos x}$

Given: ΔBCA ~ ΔSTR

Since , corresponding angles of two similar triangles are equal.

So, ∠C = ∠T ...(i) [Middle letter]

In triangle STR

$\sin T=\dfrac{\text{Side opposite to T}}{\text{Hypotenuse}}\\\\=\dfrac{26.4}{30.9}\approx0.85\\\\\cos x =\dfrac{\text{Side adjacent to T}}{\text{Hypotenuse}}\\\\=\dfrac{16.2}{30.9}\approx0.52\\\\\tan T=\dfrac{\sin T}{\cos T}\\\\=\dfrac{0.85}{0.52}\approx1.63$ ...(ii)

From (i) and (ii), we have

$\sin C\approx0.85\\\\\cos C \approx0.52\\\\\tan C\approx1.63$

4 0

3 years ago

Power Series Differential equation

KatRina [158]

The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for $y$

$\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0$

which indeed gives the recurrence you found,

$a_{n+3}=-\dfrac{n-3}{n+3}a_n$

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that $a_2=0$ , and substituting this into the recurrence, you find that $a_2=a_5=a_8=\cdots=a_{3k-1}=0$ for all $k\ge1$ .

Next, the linear term tells you that $6a_0+6a_3=0$ , or $a_3=a_0$ .

Now, if $a_0$ is the first term in the sequence, then by the recurrence you have

$a_3=a_0$
$a_6=-\dfrac{3-3}{3+3}a_3=0$
$a_9=-\dfrac{6-3}{6+3}a_6=0$

and so on, such that $a_{3k}=0$ for all $k\ge2$ .

Finally, the quadratic term gives $6a_1-12a_4=0$ , or $a_4=\dfrac12a_1$ . Then by the recurrence,

$a_4=\dfrac12a_1$
$a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1$
$a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1$
$a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1$

and so on, such that

$a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}$

for all $k\ge2$ .

Now, the solution was proposed to be

$y=\displaystyle\sum_{n\ge0}a_nx^n$

so the general solution would be

$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots$
$y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)$
$y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)$

4 0

3 years ago

I WANT SOLUTION AND EXPLANATION ANSWER IS GIVEN AT LAST!!!

Serga [27]

let the numbers be x and y

x + y = 1600 ----*

12%x = 20%y

0.12x = 0.2y

y = 0.12x

0.2

y = 0.6x

substitute in * eqn

x + 0.6x = 1600

1.6x = 1600

x = 1000

then,

y = 0.6*1000

y = 600

the numbers are 1000 and 600

6 0

2 years ago

Read 2 more answers

What is the least common multiple of 96 144 and 224

Gennadij [26K]

The least common multiple of 96, 144, and 224 is 2016.

Hope this helps! :D

5 0

3 years ago

Read 2 more answers