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3 years ago

X^2+6x+9=1 Solve each equation.

Mathematics

2 answers:

devlian [24]3 years ago

6 0

It is -4 and -2 for X

wel3 years ago

6 0

The answer is x= -2,-4

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1/2x +1/3 y = 3

LekaFEV [45]

C is the correct answer.
If you want to check it just divide both sides by six.
3/6x + 2/6y = 18/6
1/2 + 1/3y = 3

7 0

4 years ago

7:5 is equivalent to 21: ?

Svet_ta [14]

Answer:

7:5 is the same as 21: 15

Step-by-step explanation:

21/7 = 3 and 3*5=15

i divided 21 by 7 to see how many times it was multiplied to get there, so i could do the same thing to the 5

I hope this helps :)

4 0

3 years ago

Read 2 more answers

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

2log(x-3)+1=5 can someone explain how to solve for x?

Brums [2.3K]

$\bf 2log(x-3)+1=5\iff 2log_{10}(x-3)+1=5\\\\\\ 2log_{10}(x-3)=4
\\\\\\
log_{10}(x-3)=\cfrac{4}{2}\implies log_{10}(x-3)=2\\\\
-----------------------------\\\\
{{ a}}^{log_{{ a}}x}=x\impliedby \textit{using this cancellation rule}\\\\
-----------------------------\\\\
10^{\cfrac{}{}log_{10}(x-3)}=10^2\implies x-3=10^2\implies x=100+3$

5 0

4 years ago

The sum of two numbers is 40. One number is 4 less than the other. Find the numbers

yawa3891 [41]

One number is 18 and the other is 22

7 0

3 years ago