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Oksi-84 [34.3K]

2 years ago

13

Someone please solve this

1 answer:

Wittaler [7]2 years ago

8 0

Step-by-step explanation:

$2 x \times \frac{1}{4} - 1 \times \frac{1}{4} - 2 \frac{1}{4} = 1 \times \frac{1}{4}$

$\frac{2}{8} - \frac{1}{4} - \frac{2}{8} = \frac{1}{4}$

and then minus them

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How does using a model help me multiply decimals by whole numbers?

slavikrds [6]

It would help find the answer and when your done see how you got your answer
:)

6 0

3 years ago

The formula for the area of a circle is A =pie r 2. Classify the area of a circle with a radius of 0.75 centimeters as either ra

MA_775_DIABLO [31]

It is irrational.

A = π r^2 = π[3/4 cm]^2 = 9π/16 cm^2

An irrational number (Pi in this case) multiplied by a rational number (9/4) gives an irrational number.

4 0

4 years ago

Read 2 more answers

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

5 days = ? hours <br> Can you please show work so i can understand better?

son4ous [18]

24 Hours in one Day

5 Days

24 <span>× 5 = 120 Hours

Hope that helps! </span>

8 0

3 years ago

Read 2 more answers

Suppose the inside bottom of a box is painted with three colors: one-third of the bottom area is blue, one-sixth is red, and one

miss Akunina [59]

Answer:

The probability that the pebble will NOT land on color blue is 0.67.

Step-by-step explanation:

We attribute the box area as the probability of the events.

So the probability of landing in the Red Area is 1/6, the probability of landing in the Blue Area is 1/3 and the probability of landing in the Yellow Area is 1/2.

P(Red)= $\frac{1}{6}$

P(Yellow)= $\frac{1}{2}$

P(Blue)= $\frac{1}{3}$

The probability of landing in NOT Blue area is given by:

P(Not Blue)=1-P(Blue)= $1-\frac{1}{3}=\frac{2}{3} =0.67$

8 0

3 years ago