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3 years ago

11

TanA+tanB=sin(A+B)/cosAcosB

1 answer:

Elden [556K]3 years ago

4 0

Do you need to prove that it is equal?
If so:

SinA/cosA + sinB/cosB = (sinAcosB + sinBcosA)/cosAcosB

(SinAcosB + sinBcosA)/cosAcosB = (SinAcosB + sinBcosA)/cosAcosB

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Read and write the number in two other forms five hundred eight thousand

denis-greek [22]

508,000 or 5.08x10^5

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3 years ago

Julia went to the grocery store and purchased cans of soup and frozen dinners. Eack

BigorU [14]

A system of equations that could be used to determine the number of cans of soup purchased and the number of frozen dinners purchased is:

350x + 500y = 4800

y = 2x

<h3>Definition of variables </h3>

x = number of cans of soup purchased

y = number of frozen dinners purchased.

<h3>Simultaneous equations </h3>

The system of equations are known as simultaneous equations. Simultaneous equations are group of equations that have to be solved together in order to determine the required values.

<h3>Methods used to solve simultaneous equations </h3>

Elimination method
Graph method
Substitution method

To learn more about simultaneous equations, please check: brainly.com/question/25875552

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2 years ago

What is the operation used in increased by

USPshnik [31]

Adding . your increasing , so you add

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3 years ago

Read 2 more answers

74% of 643 is what number

oksano4ka [1.4K]

475.82
is the answr i think

7 0

3 years ago

Read 2 more answers

If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)

Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

$\log_dabc=\log_da+\log_db+\log_dc$

Then using the change of base formula, we can derive the relationship

$\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}$

This immediately tells us that

$\log_dc=\dfrac1{\log_cd}=\dfrac12$

Notice that none of $a,b,c,d$ can be equal to 1. This is because

$\log_1x=y\implies1^{\log_1x}=1^y\implies x=1$

for any choice of $y$ . This means we can safely do the following without worrying about division by 0.

$\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}$

so that

$\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23$

Similarly,

$\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}$

so that

$\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34$

So we end up with

$\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}$

###

Another way to do this:

$\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}$

$\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}$

$\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12$

Then

$abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}$

So we have

$\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}$

4 0

3 years ago