solve for x z= 19+2 (x+y)

We have to find value of the house in 8 years so we will plug t=8 years in given equation.

[tex] y=150000(1.004)^{\left(12t\right)} [/tex]

[tex] y=150000(1.004)^{\left(12*8\right)} [/tex]

[tex] y=150000(1.004)^{96} [/tex]

[tex] y=150000(1.46702133432) [/tex]

[tex] y=220053.200148 [/tex]

Hence house will worth approx $220053.20 in 8 years.

Heather and Joel bought a house for $157,200 and know that the house appreciates every year. They keep track of their house value for 5 years and model their data with the exponential equation y = 150000(1.004)12t How much will their house be worth in 8 years?

Solution:

Equation of the appreciation of the value of house is given by:-

[tex]y=150000(1.004)^{12t}[/tex]

We need to find the worth of the house after 8 years

So, that means time, t=8

Let us plug t=8 in the equation [tex]y=150000(1.004)^{12t}[/tex]

So, we get,

[tex]y=150000(1.004)^{12*(8)}[/tex]

[tex]y=150000(1.004)^{96}[/tex]

[tex]y=150000(1.467021)[/tex]

y=150000*1.467021

y=220053.20

In 8 years the value appreciates to $220053.20, that s, worth of house after 8 years=$22005320

The value of the investment after 15 years is approximately [tex]$4,410.67.[/tex]

To model the situation where an investment of [tex]$1,800[/tex] increases by 7% each year, we can use the exponential growth formula:

[tex]\[ A = P(1 + r)^t \][/tex]

 where:

-[tex]\( A \)[/tex]is the amount of money accumulated after n years, including interest.

- [tex]\( P \)[/tex]is the principal amount (the initial amount of money).

- [tex]\( r \)[/tex]is the annual interest rate (in decimal form).

-[tex]\( t \)[/tex] is the time the money is invested for, in years.

Given:

[tex]- \( P = \$1,800 \)[/tex]

[tex]- \( r = 7\% = 0.07 \)[/tex](as a decimal)

-[tex]\( t = 15 \)[/tex] years

 We substitute these values into the formula to get:

[tex]\[ A = 1800(1 + 0.07)^{15} \][/tex]

 Now, we calculate the value of[tex]\( A \):[/tex]

[tex]\[ A = 1800(1.07)^{15} \][/tex]

 Using a calculator, we find:

[tex]\[ A \approx 1800 \times (1.07)^{15} \approx 1800 \times 2.45037 \][/tex]

[tex]\[ A \approx \$4,410.67 \][/tex]

 Therefore, the value of the investment after 15 years is approximately [tex]$4,410.67.[/tex]

Answer:

Hence, the sum using summation notation, assuming the suggested pattern continues [tex]16+25+36+49+......+n^2+.....[/tex] is:

[tex]\sum_{n=4}^{\infty}n^2[/tex]

Step-by-step explanation:

We have to write the sum using summation notation, assuming the suggested pattern continues:

[tex]16+25+36+49+......+n^2+.....[/tex]

Clearly we may also write this pattern as:

[tex]4^2+5^2+6^2+.....+n^2+......[/tex]

So, in terms of the summand it is written as:

[tex]\sum_{n=4}^{\infty}n^2[/tex]

( We have started our summation from 4 since the term in the summation starts with 16 which is 4^2 and goes to infinity )

Answer:

Yes, the sample was biased.

Step-by-step explanation:

We are told that to choose a mascot for a new middle school, 8th-grade students were polled, and their top choice was selected as the mascot.

We know that an unbiased sample is an accurate representation of the entire population and it can help us draw conclusions about the population.

For an unbiased sample each member of a population should equally likely to be chosen and the sample should be representative of the population as a whole.  

For an unbiased sampling for mascot students should be chosen randomly from each grade of the middle school.

Since all the students were polled from the same grade, so their choice can not represent the choices of all the students of the school, therefore, the sample was a biased sample.

The middle 50% of the age of participants in the robotics competition falls between 13 and 31.

The box plot represents the age distribution of participants in a robotics competition. The box extends from the lower quartile (Q1) to the upper quartile (Q3), which in this case are 13 and 31, respectively. The line within the box represents the median, which is 16. The whiskers represent the range of the data, ending at 11 and 33.

To find the middle 50% of the data, we need to determine the values that lie between Q1 and Q3. Therefore, the middle 50% of the age of participants in the robotics competition falls between 13 and 31.

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The measure of the each part of the large circle is 90 degrees.

A circle is a bounded figure which points from its center to its circumference is equidistant. The sum of angles in a circle is 360 degrees.

Measure of the angle of each part = 1/4 x 360 = 90 degrees

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The number that will give Javier the fully simplified evaluated expression is 24.

To evaluate the expression (-8)[10 + (-5) + (-8)], we need to multiply the numbers inside the parentheses by a certain number and then add the three products together. Let's solve it step by step:

Therefore, the number that will give Javier the fully simplified evaluated expression is 24.

Answer:

A.) log 3 sqrt x(x^3+4)/2

The formula for annual compound interest, including principal sum, is:
A = P (1 + r/n) (nt)

Where:

A = the future value of the investment/loan, including interest
P = the principal investment amount (500)
r = the annual interest rate (.03)
n = the number of times that interest is compounded per year (1)
t = the number of years the money is invested (3)

A=5000(1+.03/1)^1(3)

A=546.36

Interest (I) gained is A-P

I=546.36-500

I=46.36

To answer this question, it is necessary to make use of the concepts of directional derivative and gradient of a function.

The solution is:

grad F( x, y ) ×ad(u)  =64/10

The gradient of a function f (x,y)  is a vector defined by:

grad f(x,y) = δf(x,y)/δx × i + δf(x,y)/δy × j

and directional derivative, in the direction of ab is defined by :

grad f( x,y) × Uᵃᵇ(u)   (1)    where Uᵃᵇ is a unitary vector in the direction of ab

according to that    Uᵃᵇ (u) = Uᵃᵇ/|ab|

Then if the directional derivative of f (x,y) in the direction of the vector ab is 5.

A ( 7 , 3 )   B ( 12 , 3 )    then   vector ab is:

ab = [ 12 - 7 , 3 - 3 ]     ⇒ ab = [ 5 , 0 ]  and |ab| = √ (5)² (0)²   |ab| = 5

a unitary vector in the direction of ab is:

ab(u) = 5×i/ 5 and according to equation (1)

δf(x,y) / δx × i  × 5 × i / |5| = 5

δf(x,y) / δx × i  × 5 × i = 25

δf(x,y) / δx = 5       then   f( x,y ) = 5 × x  + ????

We go on to calculate the component on j of f(x,y)

Following the same procedure

ac = ( 7 , 7 ) - ( 7 , 3 )     ⇒   ac = [ 0 ,  4 ]      |ac| = √(4)² + (0)²

|ac| = 4

Unitary vector in the direction of ac(u)  is:

ac/|ac|  = 4 × j / 4

Then :

δf(x,y) / δy × j ×  + 4 × j / 4 = 4

δf(x,y) / δy × j ×  + 4 × j  = 16

δf(x,y) / δy = - 4       and     f(x,y) =  -4×y

f(x,y) = 5 × i + 4 × j

Finally:

vector ad = [ ( 15 - 7 , 9 - 3 ) ]     ⇒  ad = ( 8 , 6 )

Unitary vector in direction ad is

ad(u) = ( 8 ×i  + 6 ×j ) / √ (8)² + (6)²     ⇒  ad(u) = ( 8 ×i  + 6 ×j ) /√ (8)² + (6)²

ad(u)  = ( 8 ×i  + 6 ×j ) /10

Now we have f (x,y ) = 5 × i + 4 × j   and ad(u) = ( 8 ×i  + 6 ×j ) /10

We can calculate the directional derivative of f(x,y) in the direction of ad with the use of equation (1)

grad F( x, y ) ×ad(u)  = ( 5 × i + 4 × j ) × ( 8 ×i  + 6 ×j ) /10

grad F( x, y ) ×ad(u)  = 4 + ( 24/10)

grad F( x, y ) ×ad(u)  =64/10

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The quotient is [tex]\(x^2 + xy + y^2\)[/tex]s [tex]\(0\)[/tex]

To divide the polynomial [tex]\(x^3 + y^3\) by \(x - y\),[/tex]division. The process yields a quotient of [tex]\(x^2 + xy + y^2\)[/tex]f [tex]\(0\)[/tex]s result indicates that the given polynomial is divisible by \(x - y\) without any remainder. The quotient represents the solution, showcasing the expression obtained when [tex]\(x^3 + y^3\)[/tex]y [tex]\(x - y\).[/tex] in the quotient is arranged in descending powers of [tex]\(x\)[/tex] ering to the instructions.

The remainder being [tex]\(0\)[/tex] irms the complete divisibility of the original polynomial by \(x - y\). This concise and ordered format aligns with the specified requirements for presenting the solution on the grid. In summary, the division of [tex]\(x^3 + y^3\) by \(x - y\)[/tex] [tex]\(x^2 + xy + y^2\)[/tex] f [tex]\(0\)[/tex]