How to write x2+y2+2x-12y-9 in Standard form

Helllppp porfavor:,)

tatuchka [14]

Answer:

1. x = 67.5

2. x = 2.5

3. x = 35.2

4. x = 2.0

5. x = 17.0

Step-by-step explanation:

Question 1

The proportion is set up in the form x/9 = 15/2. Multiply both sides by two to get rid of the two in the denominator on the right side. After doing so, multiply by 9 on both sides to get rid of the 9 in the denominator on the left:

2x/9 = 15

2x = 9(15)

Next solve for x:

2x = 135

x = 67.5

Question 2

The proportion is set up in the form 20/8.7 = 5.8/x. Multiply both sides by the second denominator, x, and then both sides by the first, 8.7. This will leave you with the work below:

20x/8.7 = 5.8

20x = 8.7(5.8)

Next, solve for x:

20x = 50.46

x = 2.523

Round to the nearest tenth:

x = 2.5

Question 3

The proportion is set up in the form 5/16 = 11/x. Multiply both sides by the second denominator, x, and then both sides by the first, 16. This will leave you with the work below:

5x/16 = 11

5x = 11(16)

Next, solve for x:

5x = 176

x = 35.2

Question 4

The proportion is set up in the form x/0.06 = 17/0.5. Multiply both sides by the second denominator, 0.5, and then both sides by the first, 0.06. This will leave you with the work below:

0.5x/0.06 = 17

0.5x = 17(0.06)

Next, solve for x:

0.5x = 1.02

x = 2.04

Round to the nearest tenth:

x = 2.0

Question 5

The proportion is set up in the form 29/x = 75/44. Multiply both sides by the second denominator, 44, and then both sides by the first, x. This will leave you with the work below:

29(44)/x = 75

29(44) = 75x

Next, solve for x:

1276 = 75x

x = 17.0133

Round to the nearest tenth:

x = 17.0

7 0

2 years ago

Triangle ABC is an isosceles triangle in which side

Shkiper50 [21]

Answer:

Option B.

Step-by-step explanation:

Consider the below figure attached with this question.

It is given that triangle ABC is an isosceles triangle.

From the below figure it is clear that the vertices of triangle are A(-2,-4), B(2,-1) and C(3,-4).

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using this formula, we get

$AB=\sqrt{(2-(-2))^2+(-1-(-4))^2}=\sqrt{16+9}=5$

$BC=\sqrt{(3-2)^2+(-4-(-1))^2}=\sqrt{1+9}=\sqrt{10}$

$AC=\sqrt{(3-(-2))^2+(-4-(-4))^2}=\sqrt{25}=5$

Now,

Perimeter of triangle ABC = AB + BC + AC

$=5+\sqrt{10}+5$

$=10+\sqrt{10}$

So, perimeter of triangle ABC is $10+\sqrt{10}$ units.

Therefore, the correct option is B.

4 0

2 years ago

Read 2 more answers

A chemist has one solution that is 20% alcohol and another that is 60% alcohol.

aleksley [76]

Answer:

solution 1= 40ml

solution 2= 160ml

Step-by-step explanation:

%alcohol= amount of alcohol/total solutionx100

0.52x200=104ml of alcohol present

0.2x L=0.2Lml of alcohol in solution one

0.6 x M=0.6Mml of alcohol in solution two

1st equation

0.2L+0.6M=104ml alcohol

times 10 to get whole

2L+6M=1040ml

2nd

same for this equation

10L+10M=2000ml

10L+10M=2000

2L+6M=1040

elimination method

=20L+20M=4000

-

=20L+60M=10400

-40M=-6400

M=160ml

2L+6 (160)=1040

2L=1040-960

2L=80

L=40ml

3 0

1 year ago

What is the x and y intercepts for 10y=5x-20

Readme [11.4K]

They represent variables

5 0

2 years ago

Read 2 more answers

(cot^2x - 1)/(csc^2x) = cos2x

Alex787 [66]

Answer:

Step-by-step explanation:

The idea here is to get the left side simplified down so it is the same as the right side. Consequently, there are 3 identities for cos(2x):

$cos(2x)=cos^2x-sin^2x$ ,

$cos(2x)=1-2sin^2x$ , and

$cos(2x)=2cos^2x-1$

We begin by rewriting the left side in terms of sin and cos, since all the identities deal with sines and cosines and no cotangents or cosecants. Rewriting gives you:

$\frac{\frac{cos^2x}{sin^2x} -\frac{sin^2x}{sin^2x} }{\frac{1}{sin^2x} }$

Notice I also wrote the 1 in terms of sin^2(x).

Now we will put the numerator of the bigger fraction over the common denominator:

$\frac{\frac{cos^2x-sin^2x}{sin^2x} }{\frac{1}{sin^2x} }$

The rule is bring up the lower fraction and flip it to multiply, so that will give us:

$\frac{cos^2x-sin^2x}{sin^2x} *\frac{sin^2x}{1}$

And canceling out the sin^2 x leaves us with just

$cos^2x-sin^2x$ which is one of our identities.

5 0

2 years ago