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erma4kov [3.2K]

3 years ago

15

How many pounds are 49 kg? And what about 50 kg? I need exact answers. Thnxs?

1 answer:

12345 [234]3 years ago

6 0

1 kg = 2.20462 lbs....so 49 kg = (49 * 2.20462) = 108.02638 lbs
and 50 kg = (50 * 2.20462) = 110.231 lbs

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Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.

Maurinko [17]

Answer:

616 square centimeters

Step-by-step explanation:

Area of a circle is equal to pi × r^2. In this case, they have asked that you substitute 3.14 or 22/7 for pi, so I will be using 22/7, instead.

Diameter is equal to 2r; to find the radius, simply divide 28 by two. This gives us a radius of 14.

Multiply 14 squared by 22/7:

(22/7)(14^2)

This gives us (22/7)(196); (4312/7); 616.

Therefore, the area of the circle is approximately 616 square centimeters.

8 0

2 years ago

Below are two parallel lines with a third line intersecting them.<br> I need help solving for x

kvv77 [185]

Answer:

its a 50 degree angle i think.

Step-by-step explanation:

its a smaller size in scale of the bigger one.

6 0

3 years ago

Read 2 more answers

P->q is false and p is true.<br> q is<br> a.true<br> b.false

kirill [66]

Answer:

Choice b. $q$ has to be false.

Step-by-step explanation:

Consider the truth table of $p \implies q$ ( $p$ implies $q$ ).

$p$ is true and $q$ is true: $p \implies q$ would be true.
$p$ is true and $q$ is false: $p \implies q$ would be false.
$p$ is false and $q$ is true: $p \implies q$ would be true.
$p$ is false and $q$ is false: $p \implies q$ would also be true.

The only combination where $p \implies q$ is false is when $p$ is true and $q$ is false. Hence, the $q\!$ here must be false.

3 0

3 years ago

Helppp pls very urgent

Ludmilka [50]

Answer:

1. 22000 millimeters

2. 1.5 grams

3. 15 centigrams

4. 5300 milliliters

5. 0.025 decagrams

6. 0.0083 decimeters

7. 0.027 decaliters

8. 28.7 millimeters

9. 5.4 decigrams

10. 1000 milligrams

Step-by-step explanation:

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

3 years ago