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3 years ago

Name the rational numbers from the list.

Mathematics

1 answer:

hram777 [196]3 years ago

3 0

The answers are A. E. I. D. H

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What is the slope of a line parallel to the line whose equation is y - x = 5?

Ghella [55]

A parallel line would have the same slope as the given line so first you'd rewrite the equation y-x=5 to y=x+5 the slope is the x value so the slope is 1.

3 0

3 years ago

Help me

Katarina [22]

The answer is 5

9+6 = 15
15/3 is 5

4 0

3 years ago

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What is the answer to this solution. -7r−4≥ 4r+2

tia_tia [17]

Answer:

The solution is:

$-7r-4\ge \:\:4r+2\quad \::\quad \:\begin{bmatrix}\mathrm{Solution:}\:&\:r\le \:\:-\frac{6}{11}\:\\ \:\:\mathrm{Decimal:}&\:r\le \:\:-0.54545\dots \:\\ \:\:\mathrm{Interval\:Notation:}&\:(-\infty \:\:,\:-\frac{6}{11}]\end{bmatrix}$

Please check the attached line graph below.

Step-by-step explanation:

Given the expression

$-7r-4\ge \:4r+2$

Add 4 to both sides

$-7r-4+4\ge \:4r+2+4$

Simplify

$-7r\ge \:4r+6$

Subtract 4r from both sides

$-7r-4r\ge \:4r+6-4r$

Simplify

$-11r\ge \:6$

Multiply both sides by -1 (reverses the inequality)

$\left(-11r\right)\left(-1\right)\le \:6\left(-1\right)$

Simplify

$11r\le \:-6$

Divide both sides by 11

$\frac{11r}{11}\le \frac{-6}{11}$

Simplify

$r\le \:-\frac{6}{11}$

Therefore, the solution is:

Please check the attached line graph below.

7 0

3 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

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Please help me with this

marishachu [46]

Answer:

Step-by-step explanation:

3 0

3 years ago

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