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3 years ago

I need help with this one, it has fractions

Mathematics

2 answers:

zepelin [54]3 years ago

8 0

Answer:

$( - 6m + 8)$

Step-by-step explanation:

$\frac{2}{3} ( - 9m + 12)$

Factor out 3 form the expression

$\frac{2}{3} \times 3( - 3m + 4)$

Reduce the numbers with greatest common divisor 3

$2( - 3m + 4)$

Distribute 2 through the parenthesis

$( - 6m + 8)$

Georgia [21]3 years ago

5 0

Answer:

$\huge\boxed{-6m+8}$

Step-by-step explanation:

$=(\frac{2}{3})(-9m+12)$

$(\frac{2}{3})(-9m)+(\frac{2}{3})(12)$

$=-6m+8$

$Thus, the~third~option ~is~correct.$

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Butoxors [25]

B=t-am is the correct answer:)

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3 years ago

A 12-foot ladder leans against the wall; the base of the ladder is 2 feet away from the wall.To the nearest tenth of a foot, how

ludmilkaskok [199]

<h3>The ladder will reach a height of 11.8 feet up the wall</h3>

Solution:

The ladder, wall and base of the ladder from wall forms a right angled triangle

Length of ladder forms the hypotenuse

Length of ladder = 12 foot

base of the ladder from wall = 2 feet

To find: height of wall

By pythagoras theorem.

$c^2 = a^2+b^2$

Where,

"c" is the Length of ladder

"a" is the base of the ladder from wall

"b" is the height of wall

Substituting the values,

$12^2 = 2^2+b^2\\\\144 = 4 + b^2\\\\b^2 = 140\\\\b = \pm 11.832 \\\\Ignore\ negative\ value\\\\b \approx 11.8$

Thus, the ladder will reach a height of 11.8 feet on wall

4 0

3 years ago

Considera um quadrado de lado 2x e um rectângulo de comprimento x e largura x  2 . Qual é o valor de x para o qual as duas figu

ki77a [65]

Answer:

3 unidades de comprimento

Step-by-step explanation:

O perímetro é definido como a soma de todos os lados.

O perímetro de um quadrado de lado 2x é dado por:

$Q=4*2x=8x$

O perímetro de um retângulo de comprimento x e largura x² é dado por:

$R=2x+2x^2$

Para que ambos os perímetros sejam iguais:

$R=Q\\8x=2x+2x^2\\x^2=\frac{6}{2}x\\ x=3$

O valor de x é igual a 3 unidades de comprimento.

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3 years ago

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astraxan [27]

Answer:

$\frac{3}{4}$

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natita [175]

Look at the attached picture...

Hope it will help you....

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2 years ago