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3 years ago

Please solve I will mark brainliest if correct

Mathematics

1 answer:

VladimirAG [237]3 years ago

6 0

Answer:

m<RQP = 62°

Step-by-step explanation:

m<RQP = m<PQB + m<RQB

m<RQP = 38° + 24°

m<RQP = 62°

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How many nanometers in 66cm

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3 years ago

What is the volume of the cone shown below?

STatiana [176]

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$V= \frac{1}{3} \pi (radius)^2h$

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3 years ago

You can model the population of a certain city between 1955-2000 by the radical function

OverLord2011 [107]

Complete Question:

You can model the population of a certain city between 1955-2000 by the radical function

P(x) = 55,000 sqrt x-1945

Using this model, in which year was the population of that city 275,000?

Answer:

In 1970 population of that city is 275,000

Solution:

Given that,

You can model the population of a certain city between 1955-2000 by the radical function :

$P(x) = 55000\sqrt{x - 1945}$

To find: year in which was the population of that city 275,000

Therefore,

x = ?

P(x) = 275000

Thus we get,

$275000 = 55000\sqrt{x - 1945}\\\\Divide\ both\ sides\ by\ 55000\\\\\frac{275000}{55000} = \sqrt{x-1945}\\\\5 = \sqrt{x - 1945}\\\\Take\ square\ on\ both\ sides\\\\25 = x - 1945\\\\x = 1945 + 25\\\\x = 1970$

Thus, in 1970 population of that city is 275,000

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2 years ago

Which description matches the transformations y=cosx undergoes to produce y=3cos(-2x)

Hoochie [10]

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3 years ago

Read 2 more answers

If the term of arthematic progression decrease in magnitude then common difference must be

Tpy6a [65]

We can say that if the term of arithmetic progression decrease in magnitude then the common difference must be negative.

An arithmetic progression is a sequence where every term is a sum of the previous term and a common difference.

If we suppose the first term of an arithmetic progression to be a, and the common difference to be d, then the arithmetic progression goes like this:

The first term, a₁ = a,

Second term, a₂ = a₁ + d = a + d,

Third term, a₃ = a₂ + d = a + 2d,

Fourth term, a₄ = a₃ + d = a + 3d, and it goes so on up to,

The n-th term, aₙ = aₙ₋₁ + d = a + (n - 1)d.

In the question, we are asked if the term of arithmetic progression decrease in magnitude then what can we say about the common difference.

As the magnitude of the term decreases, we can show that as:

The n-th term < The first term,

or, aₙ < a₁,

or, a + (n - 1)d < a,

or, a + (n - 1)d - a < a - a {Subtracting a from both sides},

or, (n - 1)d < 0.

Since, n > 1, (n - 1) > 0, thus we get the common difference, d < 0, that is, the common difference is negative.

Thus, we can say that if the term of arithmetic progression decrease in magnitude then the common difference must be negative.

Learn more about arithmetic progressions at

brainly.com/question/6561461

#SPJ4

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2 years ago