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3 years ago

PLEASE ANSWER ASAP FOR BRAINESLT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

Elza [17]3 years ago

6 0

formula is basexheight but it will need to be done in 2 separate figures so first

$10 \times 12 \times 14 = 1680$

add that to

$5 \times 12 \times 6 = 360$

equaling

<h2>2040 in cubed</h2>

hope this was helpful:)

Oksana_A [137]3 years ago

5 0

Answer:

Step-by-step explanation:

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Find the inverse of each function. h ( n ) = 4/n − 2

Darina [25.2K]

$4(x + 2) = {h(n)}^{- 1}$

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3 years ago

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Please helpppp meee !!!!

Marta_Voda [28]

Answer:

y=-9x+193

Step-by-step explanation:

Its decreasing by 9, and 175+(9x2) is 193

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3 years ago

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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

Help plz:

Mice21 [21]

Step-by-step explanation:

you will see, this is totally easy. so, once you understand that principle, I am sure you will agree with me that this costs more time putting it in here than just doing it yourself in the future.

you need to consider for addition and subtraction every variable combined with an exponent a different thing, a fruit for example.

you know exactly what to do, when you are being asked to add 3 apples to 2 apples, right ?

or, when are are asked to add 4 apples and 2 pears to 3 apples and 2 pears.

in the first case we simply get 5 (2+3) apples.

in the second case we get 7 (3+4) apples and 4 (2+2) pears.

and nothing else happens here.

we could call x² apples, y² pears, y³ plums, x kiwis, t strawberries, t² raspberries. and the constants could be another fruit or grams of sugar.

now, we simply combine the "fruits" that belong together. there is nothing more to this.

5x² + 3y³ - 4x² = x² - 3y³ (because 5-4 for x²)

y³ + 2y³ - 8 + 2 = 3y³ - 6 (1+2 for y³ and -8+2 constant)

9 - 3x - 4x + 2x - 12 = -5x - 3

-6t + 4t² - 6 + 7t - 4t² = t - 6 (4-4 for t², -6+7 for t)

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3 years ago

Solve 3x to the <br> second power <br> -2x-8=0 using an algebraic method

MatroZZZ [7]

Answer:

$x=2\textrm{ or } x=-\frac{4}{3}$

Step-by-step explanation:

Given:

The equation is given as:

$3x^{2}-2x-8=0\\3x^{2}-6x+4x-8=0\\3x(x-2)+4(x-2)=0\\(x-2)(3x+4)=0\\\\x-2=0\textrm{ or }3x+4=0\\x-2+2=0+2 \textrm{ or }3x+4-4=0-4\\x=2\textrm{ or }3x=-4\\x=2\textrm{ or}\\x=-\frac{4}{3}$

Therefore, the possible values of $x$ are 2 and $-\frac{4}{3}$

7 0

3 years ago