Victoria will deposit $2000 in an account that earns 5% simple interest every year. Her friend Corbin will deposit $1800 in an account that earns 9% interest compounded annually. The deposits are made on the same day, and no additional money will be deposited or withdrawn from the accounts. Which statement about the balances of Victoria and Corbin's accounts at the end of 3 years is true?

Answer: B

Step-by-step explanation:t+4+3−2.2t

Add 4 and 3 to get 7.

t+7−2.2t

Combine t and −2.2t to get −1.2t.

−1.2t+7

x=$4 cost of  senior citizen ticket

y= $7 cost of child ticket

Quadratic equations are the polynomial equations of degree 2 in one variable of type f(x) = ax2 + b x + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where ‘a’ is called the leading coefficient and ‘c’ is called the absolute term of f (x). The values of x satisfying the quadratic equation are the roots of the quadratic equation (α, β).

The quadratic equation will always have two roots. The nature of roots may be either real or imaginary.

First day

3x+9y=75/3

x+3y=25                                                                  -1

Second Day

8x+5y=67                                                                    -2

multiply equation 1 by -8

-8x-24y=-200

add it to equation 2

-24y+5y=-200+67

-19y=-133/-19

y= $7 cost of child ticket

plug value of y in equation1

x+21=25

x=$4 cost of  senior citizen ticket

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Answer:

\(y=-3x^{2} +6x+18\)

Step-by-step explanation:

\(y=ax^{2} +bx+c\)

Substitute the three points we have.

\(21=a+b+c\\18=4a+2b+c\\9=a-b+c\\\)

Take the first less the third.

\(2b=12\\b=6\)

\(15=a+c\\6=4a+c\\15=a+c\\\)

Take the second and subtract the first.

\(-9=3a\\a=-3\)

Therefore, we can calculate c.

\(c=18.\)

Therefore, our equation is \(y=-3x^{2} +6x+18\)

The value of the double integral ∬R (y²))/(x²+y²)dA is 18.67.

To evaluate the double integral using polar coordinates, we need to express the given region in terms of polar coordinates.

The equations of the two circles in Cartesian coordinates are:

x² + y² = 25 and x² + y² = 81

In polar coordinates, these become:

r² = 25 and r² = 81

So, the region R can be described as follows:

5 ≤ r ≤ 9 and 0 ≤ θ ≤ 2π

Now, we can change the double integral into polar coordinates:

∬R (y²)/(x²+y²)dA

= \(\int\limits^{2\pi} _0 \int\limits^9_5 (r^2sin^2\theta/(r^2) r dr d\theta\)

= \(\int\limits^{2\pi} _0 \int\limits^9_5 sin^2\theta dr d\theta\)

\(\int\limits^{2\pi} _0 \frac{1}{2} (81-25)sin^2\theta d\theta\)

= (1/2)(56)(2/3)

= 18.67

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The repeating decimal 0.8 as a geometric series 0.8 = 8/10 + (8/10)^2 + (8/10)^3 + ...

To express the repeating decimal 0.8 as a geometric series, we can start by observing the repeating pattern. In this case, the digit 8 repeats indefinitely. We can write 0.8 as follows:

0.8 = 0.8888...

To convert this into a geometric series, we need to identify a common ratio that will generate each subsequent term. In this case, the common ratio can be obtained by dividing the repeating digit by 10, which represents the shifting of the decimal point to the right. Thus, the common ratio is 8/10, which simplifies to 4/5.

Now we can express the repeating decimal 0.8 as a geometric series using the formula for an infinite geometric series:

0.8 = 8/10 + (8/10)^2 + (8/10)^3 + ...

In general, the nth term of the series is given by (8/10)^n. Since the repeating decimal has an infinite number of terms, we have successfully represented 0.8 as a geometric series.

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The 8th term of the given arithmetic sequence, which follows a constant difference of 6x + 6, is represented by the expression 37x + 51.

To find the 8th term of the arithmetic sequence, let's examine the pattern in the given sequence: -5x + 9, x + 15, 7x + 21, ...

We notice that each term in the sequence can be obtained by adding a constant difference to the previous term. To determine this constant difference, we subtract the second term from the first term and the third term from the second term:

Second term - First term:

(x + 15) - (-5x + 9) = x + 15 + 5x - 9 = 6x + 6

Third term - Second term:

(7x + 21) - (x + 15) = 7x + 21 - x - 15 = 6x + 6

We observe that the constant difference between consecutive terms is 6x + 6. Since this difference remains constant throughout the sequence, we can express the nth term of the sequence as:

nth term = (first term) + (n - 1) * (constant difference)

In our case, the first term is -5x + 9 and the constant difference is 6x + 6. Thus, the formula for the nth term becomes:

nth term = (-5x + 9) + (n - 1) * (6x + 6)

To find the 8th term, we substitute n = 8 into the formula:

8th term = (-5x + 9) + (8 - 1) * (6x + 6)

= (-5x + 9) + 7 * (6x + 6)

= -5x + 9 + 42x + 42

= 37x + 51

Therefore, the 8th term of the arithmetic sequence -5x + 9, x + 15, 7x + 21, ... is 37x + 51.

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Answer:

Connor would have $83.37 more than Henry.

Step-by-step explanation:

Connor:

Initial investment= $8,000

Interest rate= 8.2% = 0.082/12= 0.00683

Number of periods= 12*5= 60 months

Henry:

Initial investment= $8,000

Interest rate= 8% = 0.08/365= 0.00022

Number of periods= 365*5= 1,825 days

To calculate the future value, we need to use the following formula on each investment:

FV= PV*(1+i)^n

Connor:

FV= 8,000*(1.00683^60)

FV= $12,035.35

Henry:

FV= 8,000*(1.00022^1,825)

FV= $11,951.98

Difference= 12,035.35 - 11,951.98= $83.37

Connor would have $83.37 more than Henry.

Answer:

1. x < 4

2. x > 1

3. x < 5

Step-by-step explanation:

Because the sine and cosine ratios involve dividing a leg (one of the shorter two sides) by the hypotenuse (the longest side), the ratio values will never be greater than one, because (some number) / (a larger number) is always less than one.

Sine & cosine are trigonometric functions of an angle in mathematics. In the context of a right triangle, the sine and cosine of an acute angle are defined as follows: for the specified angle, the sine is the ratio of the length of the side opposite that angle to the length of the triangle's longest side (the hypotenuse), and the cosine is the ratio of the length of the adjacent leg to that of the hypotenuse.

The sine and cosine functions for an angle θ are simply denoted as sin θ and cos θ. In general, the definitions of sine and cosine can be extended to any real value expressed in terms of the lengths of specific line segments in a unit circle.

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Answer:

a. 12 inches for the shorter side

b. 1050 bushels apples

c. 7/8 cups

Step-by-step explanation:

You can solve these all by the butterfly method(aka cross multiplication)

Answer:

Area of rectangular garden = lxb = 10*5

                                             = 50 sq. inch

when area increases by six times

new area = 50*6 = 300 sq inches

now,

let x be increasement in both  length and width

(5+x) (10+x) = 300

50 + 5x + 10x + x² = 300

x² + 15x - 250 = 0

x²+ 25x-10x -250 = 0

x(x+25) - 10 (x+25) = 0

( x-10) (x+25) = 0

either x + 25 =0

x = -25 length can be negative

0r x-10 = 0

x = 10

so 10 inches is increase on both side

Step-by-step explanation:

The following can be answered by the concept of Sequence.

1. The sequence is decreasing as n increases. So, the answer is (b) decreasing.

2. The sequence is not bounded.

1. To determine whether the sequence is increasing, decreasing, or not monotonic, let's first examine the formula: an = 4n(-3). Simplifying this gives us an = -12n. Since the coefficient of n is negative, the sequence is decreasing as n increases. So, the answer is (b) decreasing.

2. To determine if the sequence is bounded, we need to see if there are upper and lower limits to the sequence. In this case, the sequence continues to decrease as n increases without any limit.

Therefore, the sequence is not bounded.

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The expressions are simplified to give;

7. 4n³(3n² + 4)

8. -3x(3x² - 4)

9. 5(k² - 8k + 2)

10. -10(6 + 6n² + 5n³)

11. 3(6n³ - 4n - 7)

12. 9(7n³ + 9n + 2)

Algebraic expressions are defined as mathematical expressions that are composed of variables, coefficients, terms, factors and constants.

These algebraic expressions are also made up of arithmetic operations, such as;

To factorize the expressions, we have;

12n⁵ + 16n³

Find the common term

4n³(3n² + 4)

-9x³ - 12x

find the common term

-3x(3x² - 4)

5k² - 40k + 10

find the common terms

5(k² - 8k + 2)

-60 + 60n² + 50n³

find the common term

-10(6 + 6n² + 5n³)

18n³ -12n - 21

find the common term

3(6n³ - 4n - 7)

63n³ + 81n + 18

Find the common term

9(7n³ + 9n + 2)

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Hazel spends (13/14) of her money in total. Therefore, the answer, in its simplest form, is 13/14.

Hazel spends (3/7) of her money in the music shop and (1/2) in the clothes shop. To find the fraction of her money she spends in total, we need to calculate the combined fraction.

Let's denote the fraction of money Hazel spends in the music shop as m and the fraction she spends in the clothes shop as c.

m = 3/7 (money spent in the music shop)

c = 1/2 (money spent in the clothes shop)

To calculate the total fraction spent, we add the fractions:

Total fraction spent = m + c

To add the fractions, we need a common denominator. The least common multiple (LCM) of 7 and 2 is 14, so we can rewrite the fractions with a common denominator:

m = (3/7) * (2/2) = 6/14

c = (1/2) * (7/7) = 7/14

Now we can add the fractions:

Total fraction spent = (6/14) + (7/14) = 13/14

Therefore, Hazel spends (13/14) of her money in total.

The answer, in its simplest form, is 13/14.

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Question

Hazel spends (3)/(7) of her money in the music shop and (1)/(2) in the clothes shop. What fraction of her money does she spend in total? Give your answer in its simplest form.

Answer:

the answer is -10 -10+5=-5

Answer:

1.50x+10=34

Step-by-step explanation:

Approximately 95% of the batteries have lifetimes between 710 hours and 990 hours.

To find the range of values that encompass 95% of the batteries, we can use the empirical rule. First, we find the z-scores corresponding to the two lifetimes of interest:

z1 = (640 - 850) / 70 = -2.43

z2 = (1060 - 850) / 70 = 2.43

Using a z-score table or calculator, we can find that the area under the standard normal distribution between -2.43 and 2.43 is approximately 0.95. Therefore, we can say that approximately 95% of the batteries have lifetimes between (850 + (-2.43) * 70) = 710 hours and (850 + 2.43 * 70) = 990 hours.

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Answer:

B.2.5

Step-by-step explanation:

The equation would be

\(-4x +8 = 2x -7\)

I believe that you would know how to solve this.

You could also sub in 2.5 for the x to check.

Good Luck!!!!!

Answer:

20/g

Step-by-step explanation:

Answer:

1/4

Step-by-step explanation:

We have 3/4 of a tank

We will use 2/3 of the  gas in the tank

That will leave us 1/4

Mathematically

3/4*2/3 = 2/4 =1/2  this is the amount that is used

Subtract this from the amount in the tank

3/4 -1/2

3/4 -2/4 = 1/4

This is the amount that is left in the tank

The mass of the original sample of phosphorus-32 was approximately 717.7 grams.

To solve this problem, we can use the radioactive decay formula:

A = Ao * 2^(-t/h)

Where:

A = resulting amount after time t

Ao = initial amount (at t=0)

t = time

h = half-life of the substance

In this case, we are given that the half-life of phosphorus-32 is 14.28 days. We want to find the initial mass, represented by Ao.

After approximately 57 days, 55 g of phosphorus-32 remain unchanged. Let's plug these values into the equation:

55 = Ao * 2^(-57/14.28)

To solve for Ao, we can isolate it by dividing both sides of the equation by 2^(-57/14.28):

55 / 2^(-57/14.28) = Ao

Using a calculator to evaluate 2^(-57/14.28), we find that it is approximately 0.07666.

Therefore, the initial mass, Ao, is:

Ao = 55 / 0.07666 ≈ 717.7 g

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The function that grows faster between (log n)^k and o(n) is o(n).

This is because the logarithmic function, (log n)^k, grows at a much slower rate than the linear function, o(n). As the value of n increases, the difference between the two functions becomes more apparent, with o(n) growing much faster than (log n)^k.

To visualize this, imagine a graph with n on the x-axis and the value of the function on the y-axis. As n increases, the value of o(n) will increase at a steady rate, while the value of (log n)^k will increase at a much slower rate.

Therefore, the function o(n) grows faster than (log n)^k.

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Answer:

no you can only use side side side when all of your triangles are equilateral triangles

Step-by-step explanation:

Answer:

diameter is twice the radius ( when given)

Answer:

5.88235294118

Step-by-step explanation:

Step 1:

46h + 56h = 600       Equation

Step 2:

102h = 600

Step 3:

h = 600 ÷ 102

Answer:

h = 5.88235294118

Hope This Helps :)

Answer:

the second one

Step-by-step explanation:

Answer:

181.818181...Miles

Step-by-step explanation:

50 = 30 + 11X

X = 181.81818181...

If you multiply 11 by 181.81 repeting you will get 20 and 30 + 20 = 50

So You would have to go 181.81 repeating miles for it to be equal

If you are rounding up it would be 182 miles to justify choosing plan B

The population of an island in 1950 was 2 million. The population grew exponentially for 63 years and reached 6.5 million in 2013. The carrying capacity of the island is estimated to be 10 million.

If the population growth rate in the future is similar to the last 50 years, what will the population be in 2050

The population is given to be increasing exponentially, which means it will follow the equation:

\($P(t) = P_0 e^{rt}$\)Here,\($P(t)$\) is the population after a period of time \($t$, $P_0$\) is the initial population, $r$ is the annual growth rate (which we are given is the same as the growth rate of the last 50 years), and \($t$\) is the time.

We can find the annual growth rate $r$ using the formula:\($$r = \frac{\ln{\frac{P(t)}{P_0}}}{t}$$\)
We know\($P_0 = 2$ million, $P(t) = 6.5$ million, and $t = 63$\) years. Substituting these values, we get:

\($r = \frac{\ln{\frac{6.5}{2}}}{63} = 0.032$\) (rounded to 3 decimal places)

Since the carrying capacity of the island is 10 million, we know that the population will not exceed this limit.

Therefore, we can use the logistic model to find the population growth over time. The logistic growth model is:

\($$\frac{dP}{dt} = r P \left(1 - \frac{P}{K}\right)$$\)

where $K$ is the carrying capacity of the environment. This can be solved to give:\($P(t) = \frac{K}{1 + A e^{-rt}}$\)

where \($A = \frac{K-P_0}{P_0}$. We know $K = 10$ million, $P_0 = 2$ million, and $r = 0.032$\). Substituting these values, we get:\($A = \frac{10-2}{2} = 4$\)

Therefore, the equation for the population of the island is:\($P(t) = \frac{10}{1 + 4 e^{-0.032t}}$\)

To find the population in 2050, we substitute\($t = 100$\) (since 63 years have already passed and we want to find the population in 2050, which is 100 years after 1950):

\($P(100) = \frac{10}{1 + 4 e^{-0.032 \times 100}} \approx \boxed{8.76}$ million\)
Therefore, the estimated population of the island in 2050, assuming constant carrying capacity and consumption per capita, is approximately 8.76 million.

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