Ask question

Ask question

All categories

3 years ago

5

What statement is always true? A the sum of 2 irrational numbers is rational B the product of 2 irrational numbers is rational C

the sum of 2 rational numbers is irrational D the product of 2 rational numbers is rational

1 answer:

Akimi4 [234]3 years ago

7 0

I think the answer is D. but I'm not sure.

You might be interested in

Which is greater 4\5 or 3/4

umka21 [38]

3/4 is greater hope that helped

7 0

4 years ago

Read 2 more answers

Please help!!! This is WAY overdue and I need to submit it by tonight or it won't be accepted!!!!

Masteriza [31]

Answer:

CCCCCCCCC

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A drawing is 12 inches wide and 10 inches long. Joe wants to enlarge each side by a scale factor of 0.3. What will be the area o

Andreyy89

Answer:

Area of the dilation = 10.8 square inches

Step-by-step explanation:

Given the original dimension

Length = 12inches

Width = 10inches

If they are dilated by a factor of 0.3

New length = 0.3 * 12 = 3.6in

New width = 0.3 * 10 = 3in

Area of the dilation = 3.6 * 3

Area of the dilation = 10.8 square inches

3 0

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

$\therefore \: x = 5$

7 0

4 years ago

Wewaii [24]

Answer:

x=5

Step-by-step explanation:

9x^2 - 2x + 25 = 8x^2 + 8x

9x^2-8x^2-2x-8x+25=0

x^2-6x+25=0 factorize

(x-5)(x-5)=0

x-5=0 then x=5

8 0

3 years ago