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3 years ago

Find 2 numbers a and b whose sum is 4 and whose difference is -2

Mathematics

1 answer:

umka21 [38]3 years ago

7 0

Answer:

a = 1, b = 3.

Step-by-step explanation:

a + b = 4

a - b = -2 Adding to eliminate b:

2a = 2

a = 1

So 1 + b = 4

giving b = 3.

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Give three numbers whose product is 9000

Klio2033 [76]

Product means, the answer of numbers being multiplied with each other.
In the given situation, we need to find three numbers that will be multiplied with each other to get a product which is 9000. Now, let’s find the possible numbers to be multiplied:
=> 300 X 30 X 1
=> 9000 x 1
=> 9000

Or this one also applies:
=> 4500 x 2 x 1
=> 9000 x 1
=> 9000 There are a lot of ways to get a product of 9000 by multiplying 3 digits.

3 0

2 years ago

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What is the maximum number of relative extrema a polynomial function can have?

creativ13 [48]

Relative extrema occur where the derivative is zero (at least for your polynomial function). So taking the derivative we get

20x3−3x2+6=0

 This is a 3rd degree equation, now if we are working with complex numbers this equation is guaranteed to have 3 solutions by the fundamental theorem of algebra. But the number of real roots are 1 which can be found out by using Descartes' rule of signs. So the maximum number of relative extrema are 1.

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3 years ago

x^2=6x+1 rewrite the equation by completing the square your equation should look like (x+a)^2=b or (x-c)^2=d

ryzh [129]

Answer:

(x - 3)² = 10

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Algebra I

Completing the Square

Step-by-step explanation:

Step 1: Define

x² = 6x + 1

Step 2: Rewrite

Subtract 6x on both sides: x² - 6x = 1
Complete the Square: x² - 6x + 9 = 1 + 9
Factor: (x - 3)² = 10

8 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:

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

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2 years ago

Larry scored 10 points less than 3 times the number of points that Ross scored. Larry scored 10 points. How many points did Ross

densk [106]

The answer is 20. you'll have to multiply 10 times 3 and get 30. Then you'll subtract the 10 from the 30 and get 20

7 0

3 years ago

Read 2 more answers

Find 2 numbers a and b whose sum is 4 and whose difference is -2​

Find 2 numbers a and b whose sum is 4 and whose difference is -2