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1 year ago

10

Slove equations with variables on both sides9 + 3f = 2f

1 answer:

Vera_Pavlovna [14]1 year ago

4 0

Given 9 + 3f = 2f

9 = 2f - 3f

9 = -f

f = -9

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An equiangular triangle has one side of length six inches. What is the height of the triangle, drawn from that side, to the near

Mandarinka [93]

9514 1404 393

Answer:

5.2 in

Step-by-step explanation:

An equiangular triangle is also equilateral. All sides will be 6 inches, and the height line will be a perpendicular bisector of the side.

Using trig, the height is ...

(3 in)tan(60°) = 3√3 in ≈ 5.2 in

Using the Pythagorean theorem, ...

h² = 6² -3² = 27

h = √27 ≈ 5.2 . . . inches

6 0

2 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

Lucy was running a 10 mile race and the graph represents her running pace for the race. If the x-axis represents miles ran and t

Ostrovityanka [42]

Answer:

C

Step-by-step explanation:

The function multiplies by 8 so 6 would be 48 and 10 would be 80.

Have a great day!

4 0

3 years ago

Drawing it out would be best can you help me please

leva [86]

Answer:

$z_1+z_2$ = (-2,5)

Step-by-step explanation:

To Find: $z_1+z_2$

Solution :

Referring the given graph

We can find the coordinates of $z_1$ and $z_2$

Coordinates of $z_1$ = (-7,1)

Coordinates of $z_2$ = (5,4)

Thus Coordinates of $z_1+z_2$ = (-7+5,1+4) = (-2,5)

Thus $z_1+z_2$ = (-2,5)

So, the point $z_1+z_2$ = (-2,5) is shown on the attached graph file

6 0

2 years ago

How do I write this expression:The quantity seven plus Z, squared

Dmitry_Shevchenko [17]

Answer:

(7+Z)²

Step-by-step explanation:

A less ambiguous way to describe the quantity might be "the square of the quantity seven plus Z".

As it is, we rely on the presence of the comma to tell us that the quantity to be squared is (7+Z). If the comma were not present, we would assume you want to add 7 to the square of Z: 7+Z².

the quantity 7 plus Z: (7+Z)

that quantity squared: (7+Z)²

6 0

3 years ago