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3 years ago

There are three people in johns family, the range of their shoe size is 4.

1 answer:

7 0

Since the range is = to max-min. The range, in this case, is 4. If there are three people and 2 wear 6, John is not one of them. So, either John's shoe size is the max or the min. If it is the max than John's size minus 6 = 4 or John's size is 10. However, it says it can't be 10. So 6 - John's size =4, which means John's size is 2. Hope this helps!

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GIVING 20 POINTS<br> If a=4 b=8 c=2 d=7 e=3 f=9 g=10<br> What does bc-c³ equal?

kozerog [31]

Answer:

Step-by-step explanation:

b=8

c=2

(8)(2)-(2)^3 = 16-8 =8

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3 years ago

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3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

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4 years ago

Matrix question help needed

kramer

The required matrix is:

$\left[\begin{array}{ccc}-25&17&0\\8&-1&3\\\end{array}\right]$

Step-by-step explanation:

We need to apply elementary row operation -2R₂+3R₁ tothe matrix:

$A=\left[\begin{array}{ccc}-3&5&2\\8&-1&3\\\end{array}\right]$

Multiplying Row 2 with -2 and Row1 with 3 and adding,

-9 15 6

-16 2 -6

----------

-25 17 0

After applying this operation, Row 1 will be changed while Row 2 will remain same because we get -2R₂+3R₁ -> R₁

The required matrix is:

$\left[\begin{array}{ccc}-25&17&0\\8&-1&3\\\end{array}\right]$

Keywords: Matrices, elementary row operation

Learn more about matrices at:

#learnwithBrainly

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3 years ago