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AleksandrR [38]
3 years ago
7

Find the coordinates of the midpoint of the segment whose endpoints are H(9, 10) and

Mathematics
2 answers:
il63 [147K]3 years ago
8 0

mid point formula: \text{coordinates of midpoint= }\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right) \\ \text{distance between two points= } \sqrt{(x_2-x_1)^2+(y_2-y_1)^2)}

Leno4ka [110]3 years ago
4 0

Answer:   The coordinate of the midpoint is m ( 8   ,   8)     and the distance HK is 4.5

Step-by-step explanation:

To find the coordinates of the point of the segment whose endpoints are H(9,10) and K(7,6), we simply use this formula;

m(x_{m} ,  y_{m})   =   m( \frac{x_{1} + x_{2}  }{2}     ,     \frac{y_{1}   +   y_{2} }{2}    )

                     

x_{1}  =  9      y_{1}  =  10          x_{2}  =   7          y_{2}   =   6

m(x_{m} ,  y_{m})     =  m  (\frac{9 + 7}{2}            ,       \frac{10  +  6}{2}     )

   

                       = m (\frac{16}{2}         ,            \frac{16}{2}          )

                          =m (  8   ,     8)

The coordinate of the midpoint is m (  8   ,     8)

To find the distance HK,

let D = distance HK

D  =   √ (x_{2}  -  x_{1} )^{2}    +    (y_{2} - y_{1})  ^{2}

       =  √(7 - 9)²   +   (6 - 10)²

       =  √(-2)²   + (-4)²

      =  √4  +  16

       

       =√20

      = 4.5  

The distance HK is 4.5

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A marketing research firm would like to survey undergraduate and graduate college students about whether or not they take out st
neonofarm [45]

Answer:

Objective Minimize 10x1 +15x2 + 12x3+18x4+15x5+ 21x6

The central total cost $ 17586 due to the number of central undergraduate students 1458 is very high.

The total minimum cost would $ 17586 +$260+ $300= $ 18146

Step-by-step explanation:

Let

X1 = # of undergraduate students from the East region,

X2 = # of graduate students from the East region,

X3 = # of undergraduate students from the Central region,

X4 = # of graduate students from the Central region,

X5 = # of undergraduate students from the West region, and

X6 = # of graduate students from the West region.

Then  the cost functions are

y1= 10x1 +15x2

y2= 12x3+18x4

y3= 15x5+ 21x6

According to the given conditions

The constraints are

x1 +x2 + x3+x4+ x5+ x6 ≥ 1500------- A

15X2 +18X4+21X6 ≥ 400---------B

21X6 ≥ 100

X6 ≥ 100/21

X6 ≥ 4.76

Taking

X6= 5

10X1 ≤ 500

X1 ≤ 500/10

X1≤ 5

18X4 ≥ 75

X4 ≥ 75/18

X4 ≥ 4.167

Taking

X4= 5

Putting the values

15X5+ 21X6 ≥ 300

15X5+ 21(5) ≥ 300

15X5+ 105 ≥ 300

15X5 ≥ 300-105

15X5 ≥ 195

X5 ≥ 195/15

X5 ≥ 13

Putting value of X6 and X4 in B

15X2 +18X4+21X6 ≥ 400

15X2 +18(5)+21(5) ≥ 400

15X2 +195 ≥ 400

15X2  ≥ 400-195

15X2  ≥ 205

X2  ≥ 205/15

X2  ≥ 13.67

Taking X2= 14

Now putting the values in the cost equations to check whether the conditions are satisfied.

y1= 10x1 +15x2

y1= 10 (5) + 15(14)= 50 + 210= $ 260

y3= 15x5+ 21x6

y3= 15 (13) + 21(5)

y3= 195+105= $ 300

x1 +x2 + x3+x4+ x5+ x6 ≥ 1500

5+14+x3+5+13+5≥ 1500

x3≥ 1500-42

x3≥ 1458

y2= 12x3+18x4

y2= 12 (1458)  + 18 (5)

y2= 17496 +90

y2= $ 17586

The cost can be minimized if the number of students from

                                 Undergraduate         Graduate

East Region              X1≤ 5                            X2  ≥ 13.67

Central                      X3≥ 1458                   X4 ≥ 4.167

West                          X5 ≥ 13                          X6 ≥ 4.76

This will result in the required number of students that is 1500

Constraints:

East Undergraduate must not be greater than 5

East Graduate must not be less than 13

Central Undergraduate must  be greater than 1458

Central Graduate must  be greater than 4

West Undergraduate must  be greater than 13

West Graduate must  be greater than 4

The central total cost $ 17586 due to the number of central undergraduate students 1458 is very high.

The east region has a least cost of $260 and west region has a cost of $300.

The total minimum cost would $ 17586 +$260+ $300= $ 18146

6 0
3 years ago
What is 7/24 written as a decimal ?
mash [69]

Pretty sure its 0.2917, not sure tho,

4 0
3 years ago
Read 2 more answers
What is the simplified value of the exponential expression 16^1\4
Natalka [10]

If you observe that 16 = 2^4, you can rewrite the expression as

16^{\frac{1}{4}} = (2^4)^{\frac{1}{4}}

Now, if you use the exponent rule (a^b)^c = a^{bc}, you may rewrite the expression again:

(2^4)^{\frac{1}{4}} = 2^{4\cdot\frac{1}{4}} = 2^1 = 2

6 0
3 years ago
Please help me. This is real confusing. ​
katrin2010 [14]

Answer:

1st problem: b) A=2500(1.01)^{12t}

2nd problem:  c) A=2500e^{.12t}

Step-by-step explanation:

1st problem:

The formula/equation you want to use is:

A=P(1+\frac{r}{n})^{nt}

where

t=number of years

A=amount he will owe in t years

P=principal (initial amount)

r=rate

n=number of times the interest is compounded per year t.

We are given:

P=2500

r=12%=.12

n=12 (since there are 12 months in a year and the interest is being compounded per month)

A=2500(1+\frac{.12}{12})^{12t}

Time to clean up the inside of the ( ).

A=2500(1+.01)^{12t}

A=2500(1.01)^{12t}

----------------------------------------------------

2nd Problem:

Compounded continuously problems use base as e.

A=Pe^{rt}

P is still the principal

r is still the rate

t is still the number of years

A is still the amount.

You are given:

P=2500

r=12%=.12

Let's plug that information in:

A=2500e^{.12t}.

6 0
3 years ago
Alice is baking a cake. She doesn’t intend to eat it, but tells you half of what you find is yours. She tells the same thing to
tankabanditka [31]

Answer: a) P = 0.5, b) P = 0.07

Step-by-step explanation:

Hi!

Lets call X₁ the time at which you arrive, and X₂ the time at which Bob arrives. Both are random variables with uniform density in the interval [0, 60] (in minutes). Their joint distribuition is uniform over the square in the image, with value P = 1/(60*60) = 1/3600.

a) For you to get more cake than Bob, you should arrive earlier. This event is A = { X₁ < X₂ }, the shaded triangle in the figure.The area of this event (set) is half the total area of the square, so P(A) = 0.5.

It makes sense, beacuse its equally probable for you or Bob to arrive earlier, as both have uniform density over the time interval.

b) In this case you arrive later than Bob, but less than 5 minutes later. So the event is B = { X₂ < X₁ < (X₂ + 5) } . This is the gray shaded area in b) part of the image. Its area is the difference two triangles (half square - blue triangle), then the probability is:

P(B) = 0.5 - \frac{(0.5*55^2)}{3600} =0.07

7 0
3 years ago
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