For which value of k is the equation below true 4,522,800,000=4.5238x10k

3) Lane and Zack purchased a house for $245,000 in 2016. The house loses 4% of its value each year,

alina1380 [7]

Answer:

$138,345

Step-by-step explanation:

This is a compound decline problem, which will be solve by the compound formula:

$F=P(1-r)^t$

Where

F is the future value (value of house at 2030, 14 years from 2016)

P is the present value ($245,000)

r is the rate of decline, in decimal (r = 4% = 4/100 = 0.04)

t is the time in years (2016 to 2030 is 14 years, so t = 14)

We substitute the known values and find F:

$F=P(1-r)^t\\F=245,000(1-0.04)^{14}\\F=245,000(0.96)^{14}F=138,344.96$

Rounding it up, it will be worth around $138,345 at 2030

8 0

3 years ago

HELP AGAIN PLEASE I TRIED IT ON MY OWN AND FAILED :(

Alex787 [66]

Answer: ∠ J = 62° , ∠ K = 59° , ∠ L = 59°

Step-by-step explanation:

It is given that it is an Isosceles Triangle, where L J ≅ K J

It follows that ∠ K ≅ ∠ L

⇒ 5x + 24 = 4x + 31

⇒ x + 24 = 31

⇒ x = 7

Input the x-value into either equation to solve for ∠ K & ∠ L:

∠ K = 5x + 24

= 5(7) + 24

= 35 + 24

= 59

∠ K ≅ ∠ L ⇒ ∠ L = 59

Next, find the value of ∠ J:

∠ J + ∠ K + ∠ L = 180 Triangle Sum Theorem

∠ J + 59 + 59 = 180

∠ J + 118 = 180

∠ J = 62

8 0

3 years ago

Using the formula T=7V+W, find the value of V when T = 26 and W = 5

ivann1987 [24]

26=7v + 5
21=7v
V=21/3
V=3

5 0

3 years ago

Read 2 more answers

Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

loris [4]

Answer:

$m=\dfrac{3}{2}$

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

$\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}$

1) Finding $(\overline{x},\overline{y})$

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

2) Finding the line through $(\overline{x},\overline{y})$ with slope m.

Given a point and a slope, the equation of a line can be found using:

$(y-y_1)=m(x-x_1)$

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

3) Find the squared vertical distances between this line and the three points.

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

4) Find the value of m that makes R minimum.

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$

$\dfrac{dR}{dm}=2(m-1)+0+2(2-m)(-1)$

now to find the minimum value we'll just use a condition that $\dfrac{dR}{dm}=0$

$0=2(m-1)+2(2-m)(-1)$

now solve for m:

$0=2m-2-4+2m$

$m=\dfrac{3}{2}$

This is the value of m for which the sum of the squared vertical distances from the points and the line is small as possible!

5 0

3 years ago

Need some help with this one.

Damm [24]

Answer:

C. 3 cans

Step-by-step explanation:

9 1/9 times 3 is 27.33 and if one can cover 10 square meters then that means you'll need 3 cans.

5 0

3 years ago