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hram777 [196]
2 years ago
13

The endpoints of EF¯¯¯¯¯ are located at E (−3, 2) and F (3, 8). Point D divides EF¯¯¯¯¯ such that ED : DF is equal to 1 : 3. For

each box, select the values in the dropdown that make the statement true. The x-coordinate of point D is and the y-coordinate of point D is
Mathematics
1 answer:
arlik [135]2 years ago
3 0

Answer:

D(x,y) = (-1.5,3.5)

Step-by-step explanation:

Given

E = (-3,2)

F = (3,8)

ED: DF = 1 : 3

Required

Determine the coordinates of D

This is calculated as thus:

D(x,y) = (\frac{mx_2 + nx_1}{m+n},\frac{my_2 + ny_1}{m+n})

Where:

(x_1,y_1) = (-3,2)

(x_2,y_2) = (3,8)

m:n = 1:3

D(x,y) = (\frac{mx_2 + nx_1}{m+n},\frac{my_2 + ny_1}{m+n})

D(x,y) = (\frac{1 * 3 + 3 *-3}{1+3},\frac{1 * 8 + 3 * 2}{1+3})

D(x,y) = (\frac{-6}{4},\frac{14}{4})

D(x,y) = (-1.5,3.5)

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The cost of 7 apples is $2.10. What is the constant of proportionality that relates to the number of apples ,x, to the cost in d
Tomtit [17]

Answer:

y is the number of apples.

Step-by-step explanation: Because if x is the cost for one apple then y is for the number of apples.

4 0
2 years ago
When the effective interest rate is 9% per annum, what is the present value of a series of 50 annual payments that start at $100
ser-zykov [4K]

Answer:

$1,109.62

Step-by-step explanation:

Let's first compute the <em>future value FV.</em>  

In order to see the rule of formation, let's see the value (in $) for the first few years

<u>End of year 0</u>

1,000

<u>End of year 1(capital + interest + new deposit)</u>

1,000*(1.09)+10  

<u>End of year 2 (capital + interest + new deposit)</u>

(1,000*(1.09)+10)*1.09 +10 =

\bf 1,000*(1.09)^2+10(1+1.09)

<u>End of year 3 (capital + interest + new deposit)</u>

\bf (1,000*(1.09)^2+10(1+1.09))(1.09)+10=\\1,000*(1.09)^3+10(1+1.09+1.09^2)

and we can see that at the end of year 50, the future value is

\bf FV=1,000*(1.09)^{50}+10(1+1.09+(1.09)^2+...+(1.09)^{49}

The sum  

\bf 1+1.09+(1.09)^2+...+(1.09)^{49}

is the <em>sum of a geometric sequence </em>with common ratio 1.09 and is equal to

\bf \frac{(1.09)^{50}-1}{1.09-1}=815.08356

and the future value is then

\bf FV=1,000*(1.09)^{50}+10*815.08356=82,508.35564

The <em>present value PV</em> is

\bf PV=\frac{FV}{(1.09)^{50}}=\frac{82508.35564}{74.35572}=1,109.616829\approx \$1,109.62

rounded to the nearest hundredth.

5 0
2 years ago
Can you explain $2000-$1433
Sophie [7]
The answer would be $567
4 0
3 years ago
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I need help with number 12
Genrish500 [490]

Answer:

150

Step-by-step explanation:

-(-4)^2 + 7(-2)^4 - 2(-3)^3

-16 + 112 - 2(-27)

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5 0
2 years ago
Read 2 more answers
Calculate the number of kilograms of mercury required to fill a pool that is 20.0 m x 50.0 m x 10.0 m. The density of mercury is
malfutka [58]

Answer:

m=135000000 Kg

Step-by-step explanation:

To find the number of kilograms of mercury we need to find how to relate density, mass and, volume. For this we shall recall the density formula:

\rho=\frac{m}{V}

where \rho is the density, m is the mass and, V is the volume.

We have the density and want to compute the mass so now we want to know the volume of the pool.

The volume of a rectangular pool is given by the fomula:

V= (length)(heigh)(width).

So for our pool

V= (20)(50)(10).

V= 10000m^{3}.

Our density is in g/mL, so the last thing we need to do before computing the mass is to express the density in Kg/m^{3} (this is because we want our mass in Kg and, we have our volume in m^{3}).

For the density conversion we have to remember that

1000mL = 1L

1L=0.001m^{3}

1000g=1Kg

so

\rho=13.5\frac{g}{mL}*\frac{1000mL}{1L}*\frac{1L}{0.001m^{3}}*\frac{1Kg}{1000g}   =13500\frac{Kg}{m^{3}}.

With this we can finally compute mass:

\rho=\frac{m}{V}

m=\rho*V

m=(13500*10000) Kg

m=135000000 Kg.

4 0
3 years ago
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