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3 years ago

4 times the sum of a number and -6 equaled 10. what's the #?

Mathematics

1 answer:

kakasveta [241]3 years ago

5 0

The answer is 8.5. you simply take the worded equation above and turn it into a math problem (a+-6) x 4

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Find the MIDPOINT between the two points.*<br> (-1,-5) and (-5, 9)

ch4aika [34]

Answer:

(-3, 2)

Step-by-step explanation:

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3 years ago

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Identify each as an equation or and expression.

olya-2409 [2.1K]

Answer:

here lets make it easy for you

An expression is a problem WITHOUT an Equal sign

An equation is a problem WITH an Equal sign

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3 years ago

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Find the inverse Laplace transforms, as a function of x, of the following functions:

inn [45]

Answer: The required answer is

$f(x)=e^x+\cos x+\sin x.$

Step-by-step explanation: We are given to find the inverse Laplace transform of the following function as a function of x :

$F(s)=\dfrac{2s^2}{(s-1)(s^2+1)}.$

We will be using the following formulas of inverse Laplace transform :

$(i)~L^{-1}\{\dfrac{1}{s-a}\}=e^{ax},\\\\\\(ii)~L^{-1}\{\dfrac{s}{s^2+a^2}\}=\cos ax,\\\\\\(iii)~L^{-1}\{\dfrac{1}{s^2+a^2}\}=\dfrac{1}{a}\sin ax.$

By partial fractions, we have

$\dfrac{s^2}{(s-1)(s^2+1)}=\dfrac{A}{s-1}+\dfrac{Bs+C}{s^2+1},$

where A, B and C are constants.

Multiplying both sides of the above equation by the denominator of the left hand side, we get

$2s^2=A(s^2+1)+(Bs+C)(s-1).$

If s = 1, we get

$2\times 1=A(1+1)\\\\\Rightarrow A=1.$

Also,

$2s^2=A(s^2+1)+(Bs^2-Bs+Cs-C)\\\\\Rightarrow 2s^2=(A+B)s^2+(-B+C)s+(A-C).$

Comparing the coefficients of x² and 1, we get

$A+B=2\\\\\Rightarrow B=2-1=1,\\\\\\A-C=0\\\\\Rightarrow C=A=1.$

So, we can write

$\dfrac{2s^2}{(s-1)(s^2+1)}=\dfrac{1}{s-1}+\dfrac{s+1}{s^2+1}\\\\\\\Rightarrow \dfrac{2s^2}{(s-1)(s^2+1)}=\dfrac{1}{s-1}+\dfrac{s}{s^2+1}+\dfrac{1}{s^2+1}.$

Taking inverse Laplace transform on both sides of the above, we get

$L^{-1}\{\dfrac{2s^2}{(s-1)(s^2+1)}\}=L^{-1}\{\dfrac{1}{s-1}\}+L^{-1}\{\dfrac{s}{s^2+1}+\dfrac{1}{s^2+1}\}\\\\\\\Rightarrow f(x)=e^{1\times x}+\cos (1\times x)+\dfrac{1}{1}\sin(1\times x)\\\\\\\Rightarrow f(x)=e^x+\cos x+\sin x.$

Thus, the required answer is

$f(x)=e^x+\cos x+\sin x.$

4 0

4 years ago

Sin(cos^-1 x)<br><br> how to solve?

tatiyna

Answer:

$sin(cos^{-1}x)=\sqrt{1-x^2}$

Step-by-step explanation:

let us assume

$cos^{-1} x=\theta\\x=cos\theta\\$

now

$sin(cos^{-1}x)$ can be written as $sin\theta$

we know $sin^2\theta+cos^2\theta=1$

so $sin^2\theta=1-cos^2\theta$

$sin\theta=\sqrt{1-cos^2\theta}$

now from above equation we can say that

$sin\theta=\sqrt{1-x^2}$

$sin(cos^{-1}x)=\sqrt{1-x^2}$

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3 years ago

Please give me the correct answer. Thanks.

lorasvet [3.4K]

Answer:

he would have to work 43 hours

Step-by-step explanation:

So if we know that at job one he gets payed 860 dollars a week and in job two he gets payed 20 per hour all we need to do is divide 860 by 20 to find the number of hours in a week and it would be 43

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3 years ago

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