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3 years ago

Ebony defines an obtuse angle as an angle that is not acute. Is Ebony’s definition valid?

Mathematics

2 answers:

Murljashka [212]3 years ago

8 0

Answer:

no because there are also right angles. an obtuse angle can best be defined as an angle greater than 90 degrees and less than 180 degrees

tekilochka [14]3 years ago

4 0

Answer:

Ebony's definition is not valid.

Step-by-step explanation:

Obtuse angle is an angle that measures more than 90 degrees but less than 180 degrees.

And acute angles measure more than zero degrees and less than 90 degrees.

We know that a right triangle is neither acute nor obtuse.

As Ebony has not mentioned right angles, so her definition is not valid.

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8/15-3/15 what’s the answer

kolbaska11 [484]

Answer:

5/15

Step-by-step explanation:

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3 years ago

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Solve y=34x y=-54x+52

jeyben [28]

Answer:

x= 13/22

Step-by-step explanation:

you can substitute y= 34 x on the second equation

34x = -54x +52 add 54x to both sides

34x +54x = -54x +54x +52

88x = 52 divide both side by 4

22x = 13 divide both sides by 22

x= 13/22

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2 years ago

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Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

2 years ago

Pls help WILL GIVE BRAINLIST

myrzilka [38]

Answer:

135.94 feet

Step-by-step explanation:

Split the problem into two parts :

Distance between viewer and building
Height of the top part of the building

Distance between viewer and building

The lower part of the building is 40 feet, as it is mentioned the viewer is 40 feet above street level
Let the distance be called 'd'
Therefore, tan37° = 40/d
tan37° = 3/4 = 0.75
⇒ 40/d = 0.75
⇒ d = 40/(3/4) = 40 x 4/3 = 160/3 = 53.3 feet

Height of the top part of the building

Let the height of the top part be 'h'
Therefore, tan61° = h/d = h/53.3
tan61° = 1.8
⇒ h/53.3 = 1.8
⇒ h = 53.3 x 1.8 = 95.94 feet

Total height of building

Lower part + Top part
40 + 95.94
135.94 feet

3 0

2 years ago

Eduardo bought a refrigerator with a sticker price of $2400. If he paid $35 a week for two years, what was the approximate marku

Norma-Jean [14]

There are 52 weeks a year
In 2 years there are 104 weeks

Total paid
35×104=3,640

markup rate
((3,640−2,400)÷2,400)×100=51.7%

It's b

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2 years ago

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