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3241004551 [841]
3 years ago
8

1. The rent of a condo unit was ₱1,200.00 for one night, to be shared equally by a group of female friends, who will be having t

heir get together in the condo unit. At the last minute, four people decided not to go to the party, thus raising the cost to each girl by ₱400.00. How many girls attended the get-together?
(With Solutions Please)
Mathematics
1 answer:
zhuklara [117]3 years ago
4 0

Answer:

6 girls were there are first

Step-by-step explanation:

use trial and error. divide 1200 by a random number like 10, and see how much they each had to pay. If we guessed 10, then each girl would hypothetically pay 120 pounds. If four girls left, then 1200/6 would be 200 pounds. there was only a 60 pound increase here, not a 400 pound one.

I checked 1200/6 which equals 200 pounds per girl. If 4 girls left, then it would be 600 pounds for the 2 remaining girls. the rent increased from 200 to 600 which is a 400 pound increase.

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The radii of two circles are in the ratio of 3 to 1. Find the area of the smaller circle if the area of the larger circle is 27
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ANSWER
3\pi sq.\: in.

EXPLANATION

Let R be the radius of the bigger circle and r, be the radius of the smaller circle.

Then their ratio is given as,

R:r=3:1

We can rewrite it as fractions to get,

\frac{R}{r} = \frac{3}{1}

We make R the subject to get,

R = 3r

The area of the bigger circle can be found using the formula,

Area=\pi {r}^{2}

This implies that,

Area=\pi ({3r})^{2}

Area=9\pi {r}^{2}

But it was given in the question that, the area of the bigger circle is 27π.

27\pi=9\pi {r}^{2}

We divide through by 9π to get,

3 = {r}^{2}

This means that,
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The area of the smaller circle is therefore

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3 years ago
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navik [9.2K]

The intervals that satisfy the given trigonometric Inequality are; 0 ≤ x < 3π/2 and 3π/2 < x ≤ 2π

<h3>How to solve trigonometric inequality?</h3>

We are given the trigonometric Inequality;

2 sin(x) + 3 > sin²(x)

Rearranging gives us;

sin²(x) - 2 sin(x) - 3 < 0

Factorizing this gives us;

(sin(x) - 3)(sin(x) + 1) < 0

Thus;

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sin(x) = 3 or sin(x) = -1

sin(x) = 3 is not possible because sin(x) ≤ 1.

Thus, we will work with;

sin(x) = -1 for the interval 0 ≤ x ≤ 2π radians.

Then, x = sin⁻¹(-1)

x = 3π/2.

Now, if we split up the solution domain into two intervals, we have;

from 0 ≤ x < 3π/2, at x = 0. Then;

sin²(0) - 2 sin(0) - 3

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Thus, the interval 0 ≤ x < 3π/2 is true.

From 3π/2 < x ≤ 2π, take x = 2π. Then;

sin²(2π) - 2 sin(2π) - 3

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Thus, the interval 3π/2 < x ≤ 2π is also true.

Read more about trigonometric inequality at; brainly.com/question/27862380

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What is the radius for the circle given by the equation x^2+(y -3)^2 = 21
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Answer:

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Step-by-step explanation:

yes i changed it because i don't want to confuse anyone:)

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