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3 years ago

Which number lines have points that represent additive inverses? Check all

Mathematics

1 answer:

frozen [14]3 years ago

4 0

Answer:

2nd and fourth one

Step-by-step explanation:

I just did it and if you look at both of them and if you look closely they each have a negative and a positive

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Find the volume of a cone with a height of 12 ft and a base diameter of 10 ft .

Tju [1.3M]

I think you know this but just in case you dont the radius is half of the diameter. That's why I put 5 instead of 10. :)

8 0

4 years ago

The angle of elevation to a nearby tree from a point on the ground is measured to be

raketka [301]

Answer: the tree is 49.25

Step-by-step explanation:

8 0

3 years ago

Suppose the distribution of home sales prices has mean k300, 000 and standard

777dan777 [17]

<u>Given</u>:

Mean,

$\mu = 300,000$

Standard deviation,

$\sigma = 50,000$

The Chebyshev theorem will be used for the solution of the given query,

i.e., $P(|X - \mu| < k \sigma ) \geq 1-\frac{1}{k^2}$

(i)

⇒ $1-\frac{1}{k^2} = 0.75$

$k = 2$

The lower limit will be:

= $\mu - k \sigma$

= $300000-2\times 50000$

= $200000$

The upper limit will be:

= $\mu + k \sigma$

= $300000+2\times 50000$

= $400000$

hence,

The 75% houses sold are between:

⇒ (200000, 400000)

(ii)

⇒ $\mu -k \sigma = Lower \ limit$

By putting the values, we get

$300000-50000k = 150000$

$300000-150000=50000k$

$150000=50000k$

$k=\frac{150000}{50000}$

$=3$

now,

⇒ $1-\frac{1}{k^2} = 1-\frac{1}{3^2}$

$=\frac{8}{9}$

$=0.8888$

$=88.88$ (%)

hence,

88.88% houses sold between 150000 and 450000.

(iii)

⇒ $\mu - k \sigma = Lower \ limit$

By putting the values, we get

$300000-50000k = 170000$

$300000-170000=50000 k$

$130000= 50000 k$

$k = \frac{130000}{50000}$

$=2.6$

now,

⇒ $1-\frac{1}{k^2} = 1-\frac{1}{2.6^2}$

$=0.8521$

$=85.21$ (%)

hence,

85.21% houses are sold between 170000 and 430000.

Learn more:

brainly.com/question/20897803

3 0

3 years ago

Please help please and thank you

Katyanochek1 [597]

Step-by-step explanation:

$\text{We know:}\\\\\sin(-\alpha)=-\sin\alpha\\\\\sin(\alpha\pm k\cdot360^o)=\sin\alpha\\\\===============================\\\\\sin(360^o-\theta)=\sin\bigg(-(\theta-360^o)\bigg)=-\sin(\theta-360^o)=-\sin\theta$

3 0

3 years ago

Inverse laplace of L^-1 {5s/s^2 + 3s - 4}

Salsk061 [2.6K]

Answer:

$e^t+e^{-4t}$

Step-by-step explanation:

We have to simplify the original function using partial fraction, hence:

$\frac{5s}{s^2+3s-4} =\frac{5s}{(s-1)(s+4)}\\\\=\frac{A}{s-1}+\frac{B}{s+4}\\\\Therefore:\\\\\frac{A(s+4)+B(s-1)}{(s-1)(s+4)}=\frac{5s}{(s-1)(s+4)}\\\\Eliminating\ the \ denominator:\\\\A(s+4)+B(s-1)=5s\\\\substitute\ s=1:\\\\A(1+4)+B(1-1)=5(1)\\\\5A=5\\\\A=1\\\\tsubstitute\ s=-4:\\\\A(-4+4)+B(-4-1)=5(-4)\\\\-5B=-20\\\\B=4\\\\Therefore\ substituting\ A\ and\ B\ gives:\\\\\frac{5s}{s^2+3s-4}=\frac{1}{s-1}+ \frac{4}{s+4}\\\\$

$From\ Laplace\ inverse:\\\\But\ L^{-1}[\frac{1}{s-a} ]=e^{at}\\\\Hence:\\\\L^{-1} [\frac{5s}{s^2+3s-4}]=L^{-1}[\frac{1}{s-1} ]+L^{-1}[\frac{4}{s+4} ]=e^{t}+4e^{-4t}\\\\L^{-1} [\frac{5s}{s^2+3s-4}]=e^{t}+4e^{-4t}$

5 0

3 years ago