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GuDViN [60]
3 years ago
9

Need help in geometry!!!

Mathematics
1 answer:
Ludmilka [50]3 years ago
7 0
Given:
JKLM: J(8,4) ; K(4,10) ; L(12,12) ; M(14,10) 

I'll just use the first vertex of every quadrilateral in comparison with J(8,4)

A(-8,-4) ; B(-4,-10) ; C(-12,-12) ; D(-14,-10) → a sequence of reflections across the x and y-axes, in any order. This is easy because each vertex has the same number but different signs. 

E(11,2) ; F(7,8) ; G(15,10) ; H(17,8) → a translation of 3 units left and 2 units up. As you can envision with a graph, from 11 to 8, you need to move to the left of the x-axis. from 2 to 4, you need to move up of the y-axis.

O(6,7) ; P(2,13) ; Q(10,15) ; R(12,13) → a translation 2 units right and 3 units down. From 6 to 8 of the x-axis and from 7 to 4 of the y axis. 

S(4,16) ; T(10,20) ; U(12,12) ; V(10,10) → This one has no pair but it seems as though this is rotated along the point U.

W(11,7) ; X(7,13) ; Y(15,15) ; Z(17,13) → a translation 3 units down and 3 units left. 
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4 - 2(x-4)/ 3= 3(2x+5)/2<br>pls answer fast​
adell [148]

Answer:

-21/22

Step-by-step explanation:

1.Apply fraction cross multiply

2. Simplify

3.Expand

4.Subtract 24 from both sides

5.simplify

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7.Simpiify, then divide both sides by -22to get answer.

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In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Interne
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Answer:

b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.

Step-by-step explanation:

Given that in a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week. The mean of the sample observations was 12.9 hours.

H_0: \bar x =12.7\\H_a: \bar x >12,7

(Right tailed test at 5% level)

Mean difference = 0.2

Std error = \frac{6}{\sqrt{1000} } \\=0.1897

Z statistic = 1.0540

p value = 0.145941

since p >alpha we do not reject H0.

b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.

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3 years ago
A standard oil drum is a cylinder with a volume of approximately 218.276 liters (or 0.218276 cubic meters). The cost of producin
Arte-miy333 [17]

Answer:

r = 2,85 feet

h = 8,56 feet

Step-by-step explanation:

Area of two circular caps (m ^(2)) : 2* pi* r^(2) (where  r is the radius of circular cap)     Area of wall (m ^(2))  is 2*pi*r*h (where h is the height of drum)

V= pi*r^(2)*h           so  h=V/pi*r^(2)   (1)

Cost in $:

Cost of wall 25 $/m^(2) plus 10% for manufacturing so:

C(w) = 27,5*2*pi*r*h $/square feet

Cost of (both top and bottom cap) = 2*40,97*pi* r^(2) so:

C(caps) = 257,29* r^(2) $/squaer feet

Total cost of cylnder = cost of wall + cost of caps

Total cost f cylinder: F(c)= 55*pi*r*h+257,29* r^(2)

F(c) = 172,7*r*h + 257,29*r^(2)         F(c)  is F(r)

Since from (1) we have h=218,276/pi*r^(2)  

F(r) = (172,7)*r*(69,514)/r^(2) +257,29*r^(2)           F(c) = 12005/r + 257,29*r^(2)

Taken the first dervative

F´(r)= -12005*(1)/r^(2) +2*257,29*r         F´(r) =  -12005*(1)/r^(2) +514,58*r

f F´(r) = 0          -12005*(1)/r^(2) + 514,58*r =0

-120005 + 514,58*^(3) = 0          514,58*^(3) = 12005      r =cubic root (23,32)

r = 2,85 ft

if we replace this value en F(r) we can see F(r) tend to infinite both when r tend to 0 and when r tends to infnite so there is a minimun for the functon

r = 2,85 ft    and   h = 8,56 ft

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