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Anna71 [15]
2 years ago
8

Please help on the air force question guys

Mathematics
1 answer:
Elza [17]2 years ago
4 0
I answer . The white ones are better trust me, but u have to keep them keep clean!
You might be interested in
What is the slope of the line y = -2?
Verdich [7]

Answer:

0

Step-by-step explanation:

There is no x, therefore there is no slope.

4 0
3 years ago
A room can be cleaned in 2 hours if rob and Harry work together. Ron could clean it in 3 hours. How long can harry clean it?
tresset_1 [31]

Answer:

6 hours

Step-by-step explanation:

Let the total work be X

Combined speed: X/2

Ron's speed: X/3

Harry's speed: S

X/2 = X/3 + S

S = X/2 - X/3 = (3X - 2X)/6 = X/6

Harry's speed is X/6, so he'll take 6 hours

6 0
3 years ago
find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0​
Citrus2011 [14]

Answer:

Radius: r =\frac{\sqrt {21}}{6}

Center = (-\frac{3}{2}, -\frac{2}{3})

Step-by-step explanation:

Given

9x^2 + 9y^2 + 27x + 12y + 19 = 0

Solving (a): The radius of the circle

First, we express the equation as:

(x - h)^2 + (y - k)^2 = r^2

Where

r = radius

(h,k) =center

So, we have:

9x^2 + 9y^2 + 27x + 12y + 19 = 0

Divide through by 9

x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0

Rewrite as:

x^2  + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}

Group the expression into 2

[x^2  + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}

[x^2  + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}

Next, we complete the square on each group.

For [x^2  + 3x]

1: Divide the coefficient\ of\ x\ by\ 2

2: Take the square\ of\ the\ division

3: Add this square\ to\ both\ sides\ of\ the\ equation.

So, we have:

[x^2  + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}

[x^2  + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2

Factorize

[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2

Apply the same to y

[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}

Add the fractions

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}

[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}

Recall that:

(x - h)^2 + (y - k)^2 = r^2

By comparison:

r^2 =\frac{7}{12}

Take square roots of both sides

r =\sqrt{\frac{7}{12}}

Split

r =\frac{\sqrt 7}{\sqrt 12}

Rationalize

r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}

r =\frac{\sqrt {84}}{12}

r =\frac{\sqrt {4*21}}{12}

r =\frac{2\sqrt {21}}{12}

r =\frac{\sqrt {21}}{6}

Solving (b): The center

Recall that:

(x - h)^2 + (y - k)^2 = r^2

Where

r = radius

(h,k) =center

From:

[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}

-h = \frac{3}{2} and -k = \frac{2}{3}

Solve for h and k

h = -\frac{3}{2} and k = -\frac{2}{3}

Hence, the center is:

Center = (-\frac{3}{2}, -\frac{2}{3})

6 0
2 years ago
In circle P, EG and FH are diameters.<br> What is m EH?
zvonat [6]

Answer:

m(\widehat{EH}) = 130°

Step-by-step explanation:

Angles subtended at the center by the arcs (\widehat{HG}) and (\widehat{EF}) are ∠HPG and  ∠EPF.

Since these angles are the vertical angles both will be equal.

m∠HPG ≅ m∠EPF

3x - 10 = 2x + 10

3x - 2x = 10 + 10

x = 20

Therefore, m(\widehat{HG})=(3\times 20) - 10

                               = 50°

Similarly m(\widehat{EF}) = 50°

In the same way angles subtended at the center will be equal.

m∠EPH ≅ m∠FPG

and m(\widehat{EH})=m(\widehat{FG})

Since m(\widehat{EH})+m(\widehat{FG})+m(\widehat{EF})+m(\widehat{HG})=360°

m(\widehat{EH})+m(\widehat{EH})+50+50=360

2m(\widehat{EH})=360-100

m(\widehat{EH})=130

Therefore, measure of arc EH = 130°

3 0
3 years ago
Find area of the shape
pentagon [3]
The answer is a. You find the area of each shape and add them
4 0
3 years ago
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