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LuckyWell [14K]

2 years ago

12

If the inflation rate doubles, does the percentage decrease in buying power also double? If not, is the new percentage decrease

in buying power more than twice the old one or less than twice the old one? Interpret your answer.

1 answer:

Nadya [2.5K]2 years ago

6 0

Answer: It doubles

Step-by-step explanation:

Inflation erodes the value of a currency thereby reducing the purchasing power that customers have as they will only be able to buy less gods and services for the same amount as before.

An increase in inflation is directly proportional to a decrease in purchasing power which means that if the inflation rate doubles, the decrease in purchasing power doubles as well.

For instance, if inflation is 10%, the value of a dollar becomes 0.9c. If Inflation doubles to 20%, the dollar goes to 0.8c. Meanwhile the purchasing power would have gone from $1 to 0.8c which would translate to a 20% decrease.

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Each cone of the hourglass has a height of 15 millimeters. The total height of the sand within the top portion of the hourglass

BARSIC [14]

Calculate the volume of sand, the cone will be completely filled and the cylinder will have sand up to the 30 mm level.

Volume of sand will = the volume of the cone:
= 1/3 * pi * (6 mm)^2 * 15
= 1/3 * pi * 36 * 15 =180 pi cubic mm

the cylinder will have a volume of sand equal to:
= pi * (6 mm)^2 * 30 mm
= pi * 36 sq mm * 30 mm
= 1080 pi cubic mm

The total sand is the sum:
= 1080 pi cubic mm +180 pi cubic mm
= 1260 pi cubic mm. <span>
</span>

5 0

2 years ago

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In a necklace the ratio of yellow beads to blue beads is 3:1. What percentage of beads are yellow?

navik [9.2K]

3 yellow beads and 1 blue bead

4 0

2 years ago

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Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago

Find the eccentricity, b. identify the conic, c. give an equation of the directrix, and d. sketch the conic.

densk [106]

Answer:

a) 10/3

b) hyperbola

c) x = ± 6/5

Step-by-step explanation:

a) A conic section with a focus at the origin, a directrix of x = ±p where p is a positive real number and positive eccentricity (e) has a polar equation:

$r=\frac{ep}{1\pm e*cos\theta}$

Given the conic equation: $r=\frac{12}{3-10cos\theta}$

We have to make it to be in the form $r=\frac{ep}{1\pm e*cos\theta}$ :

$r=\frac{12}{3-10cos\theta}\\\\multiply\ both\ sides\ by\ \frac{1}{3} \\\\r=\frac{12*\frac{1}{3}}{(3-10cos\theta)*\frac{1}{3}}\\\\r=\frac{12*\frac{1}{3}}{3*\frac{1}{3}-10cos\theta*\frac{1}{3}}\\\\r=\frac{4}{1-\frac{10}{3}cos\theta } \\\\r=\frac{\frac{10}{3}(\frac{6}{5} ) }{1-\frac{10}{3}cos\theta }$

Comparing with $r=\frac{ep}{1\pm e*cos\theta}$

e = 10/3 = 3.3333, p = 6/5

b) since the eccentricity = 3.33 > 1, it is a hyperbola

c) The equation of the directrix is x = ±p = ± 6/5

6 0

2 years ago

Write each fraction as a sum of unit fractions 7/10

seropon [69]

Answer:

3/10+4/10=7/10

Step-by-step explanation:

8 0

2 years ago