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stepan [7]
2 years ago
11

One year, the population of a city was 164,000. Several years later it was 182,040. Find the percent increase.

Mathematics
2 answers:
user100 [1]2 years ago
7 0

Answer:

0

Step-by-step explanation:

ElenaW [278]2 years ago
3 0
182,040-164,000 that’s the answer YOUR WELCOME
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Augie burns 225 calories per 30 minutes of bicycling How many minutes would Augie have to bike to burn 150 calories
lianna [129]
Well first 225/30 is 7.5 so you times 7.5 by any number till you get 150, which the answer is 20 minutes to burn 150 calories

work:
223/30 = 7.5
7.5 * 20 = 150
8 0
3 years ago
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Winnie Bracken opened a savings account at Dallas Trust Bank on March 1. It pays 4% interest compounded quarterly. She opened he
fiasKO [112]

The compounded interest is applied to the amount in the account at end

of a period specified in the rate of compounding.

  • The amount in the account at the end of 6 quarters is approximately <u>$16,767.2</u>

Reasons:

The interest paid on the account = 4% compounding

Amount with which she opened the account = $10,000

Amount she makes as deposit at the end of each quarter = $1,000

Therefore;

The interest per quarter = 4% ÷ 4 = 1%

Amount in the account after the end first quarter, A₁, is therefore;

A₁ = 10,000 × 0.01 + 10,000 + 1,000 = 11,100

The amount in the second quarter, A₂, is given as follows;

A₂ = 11,100 × 0.01 + 11,100 + 1,000 = 12211

A₃ = 12211 × 0.01 + 12211 + 1,000 = 13333.11

A₄ = 13333.11 × 0.01 + 13333.11 + 1,000 = 14466.4411

A₅ = 14466.4411 × 0.01 + 14466.4411 + 1,000 = 15611.105511

At the end of the 6th quarter, we have;

A₆ = 15611.105511 × 0.01 + 15611.105511 + 1,000 = 16767.2165991

The amount in the account at the end of 6 quarters, A₆ ≈ $16,767.2

Learn more about compounding interest rate here:

brainly.com/question/8806008

3 0
2 years ago
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If PQRS is a parallelogram with two adjacent congruent sides, then _____________________.
kolezko [41]
A.) it must be a rhombus
8 0
3 years ago
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wo point charges lie on the $x$ axis. A charge of +6.24 $\mu C$ is at the origin, and a charge of -9.55 $\mu C$ is at $x$ = 12.0
Andreyy89

Answer:

E_n=34,467,075.42\ N/C

Step-by-step explanation:

<u>Electric Field</u>

The electric field produced by a point charge Q at a distance d is given by

\displaystyle E=K\cdot \frac{Q}{d^2}

Where

K = 9\cdot 10^9\ Nw.m^2/c^2

The net electric field is the vector addition of the individual electric fields produced by each charge. The direction is given by the rule: If the charge is positive, the electric field points outward, if negative, it points inward.

Let's calculate the electric fields of each charge at the given point. The first charge q_1=+6.24\mu C=6.24\cdot 10^{-6}C is at the origin. We'll calculate its electric field at the point x=-3.85 cm. The distance between the charge and the point is d=3.85 cm = 0.0385 m, and the electric field points to the left:

\displaystyle E_1=9\cdot 10^9\cdot \frac{6.24\cdot 10^{-6}}{0.0385^2}

E_1=37,888,345.42\ N/C

Similarly, for q_2=-9.55\mu C=-9.55\cdot 10^{-6}C, the distance to the point is 12 cm + 3.85 cm = 15.85 cm = 0.1585 m. The electric field points to the right:

\displaystyle E_2=9\cdot 10^9\cdot \frac{9.55\cdot 10^{-6}}{0.1585^2}

E_2=3,421,270\ N/C

Since E1 and E2 are opposite, the net field is the subtraction of both

E_n=37,888,345.42\ N/C-3,421,270\ N/C

\boxed{E_n=34,467,075.42\ N/C}

6 0
3 years ago
Performance task trigonometric identities
marishachu [46]

#1

\\ \sf\longmapsto cos75°

\\ \sf\longmapsto cos(45+30)°

\\ \sf\longmapsto cos45cos30+sin45.sin30

\\ \sf\longmapsto \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2}

\\ \sf\longmapsto \dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}}

\\ \sf\longmapsto \dfrac{\sqrt{3}+1}{2\sqrt{2}}

#2

\\ \sf\longmapsto tan\left(-\dfrac{\pi}{12}\right)

\\ \sf\longmapsto tan(-15°)

\\ \sf\longmapsto \dfrac{sin(-15)}{cos(-15)}

\\ \sf\longmapsto \dfrac{-sin15}{cos15}

\\ \sf\longmapsto -\dfrac{\dfrac{\sqrt{3}-1}{2\sqrt{2}}}{\dfrac{\sqrt{3}+1}{2\sqrt{2}}}

\\ \sf\longmapsto -\left(\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\right)

5 0
3 years ago
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