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2 years ago

11

___ cosb =1/2 sin(a+b)+sin(a-b)?

2 answers:

vodomira [7]2 years ago

8 0

Answer:

That would be sina.

Step-by-step explanation:

sin(a+b) = sinacosb + cosasinb

sin(a-b) = sinacosb - cosasinb

Adding we get sin(a+b) + sin(a-b) = 2sinaccosb

so sinacosb = 1/2sin(a+b) + sin(a-b)

dem82 [27]2 years ago

6 0

Answer:

sina

Step-by-step explanation:

We have

sin(a+b) = sina cosb + cosa sinb ---------------------eqn 1

sin(a-b) = sina cosb - cosa sinb ---------------------eqn 2

eqn 1 + eqn 2

sin(a+b) + sin(a-b) = sina cosb + cosa sinb + sina cosb - cosa sinb

sin(a+b) + sin(a-b) = 2sina cosb

$sina\times cosb =\frac{1}{2}(sin(a+b) + sin(a-b))$

So answer is sina

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Bobby's investment of $225,000 loses value at a rate of 3% per year. Use an exponential function to find the value of the invest

AleksAgata [21]

We have been given that Bobby's investment of $225,000 loses value at a rate of 3% per year. We are asked to find the value of the investment after 10 years.

We will us exponential decay function to solve our given problem.

We know that an exponential function is in form $y=a\cdot (1-r)^x$ , where,

y = Final amount,

a = Initial amount,

r = Decay rate in decimal form,

x = Time.

Let us convert 3% into decimal.

$3\%=\frac{3}{100}=0.03$

Upon substituting $a=\$225,000$ , $r=0.03$ and $x=10$ , we will get:

$y=\$225,000(1-0.03)^{10}$

$y=\$225,000(0.97)^{10}$

$y=\$225,000(0.7374241268949283)$

$y=\$165,920.4285513588675$

Upon rounding to nearest dollar, we will get;

$y\approx \$165,920$

Therefore, the value of the investment after 10 years would be $\$165,920$ .

5 0

2 years ago

Determine whether each of the following functions is a solution of Laplace's equation uxx + uyy = 0. (Select all that apply.) u

Naddika [18.5K]

Answer with Step-by-step explanation:

We are given that Laplace's equation

$u_{xx}+u_{yy}=0$

We have to determine given function is solution of given laplace's equation.

If a function is solution of given Laplace's equation then it satisfy the solution.

1. $u=e^{-x}cosy-e^{-y}cosx$

Differentiate w.r.t x

Then, we get

$u_x=-e^{-x}cosy+e^{-y}sinx$

Again differentiate w.r.t x

$u_{xx}=e^{-x}cosy+e^{-y}cosx$

Now differentiate u w.r.t y

$u_y=-e^{-x}siny+e^{-y}cosx$

Again differentiate w.r.t y

$u_{yy}=-e^{-x}cosy-e^{-y}cosx$

Substitute the values in given Laplace's equation

$e^{-x}cosy+e^{-y}cosx-e^{-x}cosy-e^{-y}cosx=0$

Hence, given function is a solution of given Laplace's equation.

2. $u=sinx coshy+cosx sinhy$

Differentiate w.r.t x

$u_x=cosx coshy-sinx sinhy$

Again differentiate w.r.t x

$u_{xx}=-sin x coshy-cosxsinhy$

Now, differentiate u w.r.t y

$u_y=sinx sinhy+cosx coshy$

Again differentiate w.r.t y

$u_{yy}=sinx coshy+cosx sinhy$

Substitute the values then we get

$-sinx coshy-cosxsinhy+sinxcoshy+cosx sinhy=0$

Hence, given function is a solution of given Laplace's equation.

4 0

2 years ago

The weekly salary paid to employees of a small company that supplies part-time laborers averages $750 with a standard deviatio

poizon [28]

Answer:

(a) The fraction of employees is 0.84.

(b)

$\mu=850\\\\\sigma=450$

(c)

$\mu=787.5\\\\\sigma=472.5$

(d) No. The left part of the distribution would be truncated too much.

Step-by-step explanation:

(a) If the weekly salaries are normally distributed, estimate the fraction of employees that make more than $300 per week.

We have to calculate the z-value and compute the probability

$z=\frac{X-\mu}{\sigma}= \frac{300-750}{450}=\frac{-450}{450}=-1\\\\P(X>300)=P(z>-1)=0.84$

(b) If every employee receives a year-end bonus that adds $100 to the paycheck in the final week, how does this change the normal model for that week?

The mean of the salaries grows $100.

$\mu_{new}=E(x+C)=E(x)+E(C)=\mu+C=750+100=850$

The standard deviation stays the same ($450)

$\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(x+C)-(\mu+C)]^2} } =\sqrt{\frac{1}{N} \sum{(x+C-\mu-C)^2} }\\\\ \sigma_{new}=\sqrt{\frac{1}{N} \sum{(x-\mu)^2} } =\sigma$

(c) If every employee receives a 5% salary increase for the next year, how does the normal model change?

The increases means a salary X is multiplied by 1.05 (1.05X)

The mean of the salaries grows 5%, to $787.5.

$\mu_{new}=E(ax)=a*E(x)=a*\mu=1.05*750=487.5$

The standard deviation increases by a 5% ($472.5)

$\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(ax)-(a\mu)]^2} } =\sqrt{\frac{1}{N} \sum{a^2(x-\mu)^2} }\\\\ \sigma_{new}=\sqrt{a^2}\sqrt{\frac{1}{N} \sum{(x-\mu)^2}}=a*\sigma=1.05*450=472.5$

(d) If the lowest salary is $300 and the median salary is $525, does a normal model appear appropriate?

No. The left part of the distribution would be truncated too much.

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2 years ago

Dan invests £2200 into a savings account. The bank gives 5% compound interest for the first 3 years and 4% thereafter. How much

Naily [24]

Present value = 2200
interest rates =5% for 3 years, and 4% afterwards
no. of periods = 7

Future value=2200*(1.05)^3(1.04)^(7-3)
=<span>£</span>2979.367 (please round to appropriate number of decimals)

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2 years ago

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Malcolm has $638.54 in his checking account. He must maintain a $500 balance to avoid a fee. He wrote a check for $159.50 today.

e-lub [12.9K]

Your answer is c, 638.54 − 159.50 + x ≥ 500; x ≥ $20.96

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2 years ago

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