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3 years ago

What is the value of 412.321 if the underlined diget is 1 on the part .321?

Mathematics

2 answers:

MArishka [77]3 years ago

6 0

The one on the part .321 is in the thousandths place

Pachacha [2.7K]3 years ago

6 0

The underlined digit is the thousandths place

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Find the distance between each pair of points.If necessary round to the nearest tenth

Dovator [93]

Answer:

Step-by-step explanation:

22) J(2,-1); K(2,5)

Distance = √(x₂-x₁)² + (y₂-y₁)²

= √( 2- 2)² + (5 - [-1] )²

= √0 + (5 + 1 )²

=√ ( 6 )² = 6

25) A(0,3) ; B (0,12)

distance =√(0-0)² + (12 - 3 )²

=√( 9)² = 9

28) Q(12,-12) ; T(5,12)

distance = √(5 - 12)² + (12 - [-12] )²

= √( -7)² + (12 + 12] )²

= √ 49 + (24)²

= √ 49 + 576

= √625

=25

3 0

3 years ago

What is 1/10 into a fraction

Katyanochek1 [597]

You just put it in a fraction 1/10

4 0

2 years ago

9y -4y+3 if y =7 solve this

BigorU [14]

Plug 7 into each y. 9(7) -4(7) +3
63-28+3
The answer is 38

5 0

2 years ago

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

The standard form of the equation of a circumference is given by the following expression:

 $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$

On the other hand, the general form is given as follows:

 $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ 

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

 $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

First. a) matches 5) 

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

Second. b) matches 1) 

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

Third. c) matches 3)

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

Fourth. d) matches 6)

6 0

3 years ago

Read 2 more answers

What are the solution

AleksandrR [38]

Answer:

Step-by-step explanation:

We can get two equations from the inequality:

$3x+2>9\\3x+2>-9$