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3 years ago

SUBJECT: inverse Trigonometric Ratios please help me asap

Mathematics

1 answer:

san4es73 [151]3 years ago

4 0

Answer:

sert

Step-by-step explanation:

Which two of the following sentences from the article include central ideas of the article?

In 43 B.C., Octavian, Antony and Marcus Aemilus Lepidus established the Second Triumvirate.

Cleopatra’s seized treasure allowed him to pay his soldiers, making them more loyal.

During his 40-year rule, Augustus nearly doubled the size of the empire.

The family tree became more tricky after Augustus had his stepson Tiberius briefly marry his daughter, and then adopted Tiberius outright as son in A.D. 4 and named him next in line to rule.Which two of the following sentences from the article include central ideas of the article?

In 43 B.C., Octavian, Antony and Marcus Aemilus Lepidus established the Second Triumvirate.

Cleopatra’s seized treasure allowed him to pay his soldiers, making them more loyal.

During his 40-year rule, Augustus nearly doubled the size of the empire.

The family tree became more tricky after Augustus had his stepson Tiberius briefly marry his daughter, and then adopted Tiberius outright as son in A.D. 4 and named him next in line to rule.

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olganol [36]

Answer:

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Step-by-step explanation:

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3 years ago

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kozerog [31]

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2 years ago

Please help me. I suck at math

serg [7]

Answer:

Answer is ...Obtuse

Step-by-step explanation:

$\frac{m$

$\frac{5x + 4}{2} = \frac{3}{2} x + 21$

$\frac{5x + 4}{2} = \frac{3x + 42}{2}$

$5x + 4 = 3x + 42$

$5x - 3x = 42 - 4$

$2x = 38$

$x = 19$

$substituting \: x = 19 \: \:in \: \: \: \: \: \: \: \\ m$

We get,

m<ABC =

$(5 \times 19) + 4 = 99$

Hence <ABC is obtuse

4 0

3 years ago

Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Ann [662]

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$ .

7 0

3 years ago

Monday’s high temperature was 6°C. The low temperature was −3°C.

Ivanshal [37]

Answer:

Step-by-step explanation:

the temperature went down it became colder

it went down 8

5 0

3 years ago