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3 years ago

5 is the same as _ and _.

Mathematics

1 answer:

nirvana33 [79]3 years ago

4 0

1/2 and 1/3 1/4. Is the answer I automatically thought of

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Aliyah bought a car for $32,000. It is now worth $5000 less than half of what she purchased it for. To the nearest tenth of a pe

Reil [10]

32,000 x 0.5 = 16000

16000- 5000 = 11000

11000 / 32000= 0.34375

1- 0.344= 0.656

Answer : D

3 0

3 years ago

A student has a savings account earning 3% simple interest. She must pay $1900 for first-semester tuition by September 1 and $19

Anastaziya [24]

Answer:

$3781.19

Step-by-step explanation:

Let us assume that the student has to earn $(1900 + x) by September 1 so that he can pay the $1900 tuition fee by September 1 and the remaining $x will grow at 3% simple interest to make him able to pay another tuition fee of $1900 by January 1.

So, we can write $x( 1 + \frac{4 \times 3}{12 \times 100}) = 1900$

{Because September 1 to January 1 is 4 months and the monthly simple interest rate is $\frac{3}{12}$ %}

⇒ 1.01x = 1900

⇒ x = $1881.19 (Rounded to the nearest cents)

Therefore, the student has to earn $(1900 + 1881.19) = $3781.19 (Answer)

8 0

3 years ago

Solve the inequality:

yarga [219]

Answer:

its A

Step-by-step explanation:

Because m > 6/7. and if you do the sloving it will be 2 or greater

Hope that helped

follow if you need more help

3 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

Find the product : 2p (4p² + 5p + 7)

Ber [7]

<h3>Answer: 8p^3 + 10p^2 + 14p</h3>

Explanation:

The outer term 2p is distributed among the three terms inside the parenthesis. We will multiply 2p by each term inside

2p times 4p^2 = 2*4*p*p^2 = 8p^3

2p times 5p = 2*5*p*p = 10p^2

2p times 7 = 2*7p = 14p

The results 8p^3, 10p^2 and 14p are added up to get the final answer shown above. We do not have any like terms to combine, so we leave it as is.

6 0

3 years ago

5 is the same as ___ and ___.

5 is the same as _ and _.