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3 years ago

I want to find the coordinates of each point of the face

Mathematics

1 answer:

Sergio039 [100]3 years ago

7 0

Answer:

A) (2,5)

B) (4,4)

C) (3,2)

D) (3,-1)

E) (1,-3)

F) (-1,-3)

G) (-3,-1)

H) (-3,2)

I) (-4,4)

J) (-2,5)

K) (-1,4)

L (1,4)

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a technology company makes more than 5 printer every hour. Which graph represent the number of printers made in 4 hours?

kiruha [24]

Answer:

See the attached file

Step-by-step explanation:

Given data

Say numbers of printer made per hour = 5 printers

Hence in 1 hour, they will make 5 printers

in 4 hours they will make

=4*5

=20 printers

The graph of this situation when plotted will give a straight line graph

Kindly find attached a straight line graph for your reference

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3 years ago

What fraction of two pounds is twenty pence?

Reptile [31]

Hey there

High light all of your key terms of this problems (fraction, of two pounds, and twenty pence)

Answer: We have $200$ pence of the $2$ $Lbs$ , which is also $20$ pence in the $\frac{1}{10} = 10%$

Good luck on your assignment and enjoy your day!

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4 years ago

This equation shows how the cost of a corporate team-building event depends on the number

sdas [7]

D=18 it would cost $18

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3 years ago

p is the point on the line 2x+y-10=0 such that the length of OP, the line segment from the origin o to p is a minimum. find the

REY [17]

Coolio

alright, so the disatnce formula is
$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

solve for y in the equation given on top
2x+y-10=0
y=10-2x
ok
so (x1,y1)=(0,0), the first point
(x2,y2)=(x,10-2x)
so subsitute to get

$D=\sqrt{(x-0)^2+(10-2x-0)^2}$
$D=\sqrt{(x)^2+(10-2x)^2}$
$D=\sqrt{x^2+4x^2-40x+100}$
$D=\sqrt{5x^2-40x+100}$
find the miniumum
we only need to minimize what is under the square root so
take the derivitive of 5x²-40x+100
that is 10x-40
it is 0 at x=4
when x<4, the derivitive is negative
when x>4, the derivitive is positve
goes from negative to positive
so it is a minimum

at x=4
find y
y=10-2x
y=10-2(4)
y=10-8
y=2

(4,2)
find min distance

$D=\sqrt{(x-0)^2+(y-0)^2}$
$D=\sqrt{(4-0)^2+(2-0)^2}$
$D=\sqrt{(4)^2+(2)^2}$
$D=\sqrt{16+4}$
$D=\sqrt{20}$
$D=2\sqrt{5}$

the point, P, is (4,2) and the minimum distance is 2√5

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3 years ago

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dem82 [27]

D is the correct answer

8 0

3 years ago