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3 years ago

Solving x+ 8 = –3 gives: (i) x = 11 (ii) x = 12 (iii) x = –11 (iv) x = –12

Mathematics

2 answers:

olchik [2.2K]3 years ago

8 0

Answer:

x+8=-3 gives

x=-11 is the required value

yan [13]3 years ago

6 0

Answer:

Step-by-step explanation:

x+8=-3

x=-3-4

so,x=-11

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Read 2 more answers

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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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

If s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n st ?

yulyashka [42]

Answer:

A) None

Step-by-step explanation:

1) $s^t$ shoudnt neccesarily be a factor of nst, for example, if s = 3, t = 4, and n = 12, then both s and t are factors of n, but $3^4 = 81$ is not a factor of nst = 144.

2) $(st)^2$ shoudnt neccesarily be a factor of nst. Let s be 4, let t be 6, and let n be 12. Then n is a factor of both s and t, but $(st)^2 = 24^2$ is not a factor of nst = 12*24. In fact, it is a greater number.

3) Again, s+t isnt necessarily a factor of nst, let s be 2 and t be 3. Then both s and t are factor of n = 12. However 5 = s+t is not a factor of nst = 72.

So, neither of the three options is guaranteed to be a factor of nst. In fact, for s = 4, t = 6, and n = 12, none of the three options are valid.

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3 years ago