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3 years ago

I REALLY NEED HELP PLEASE HELP ME :(

Mathematics

1 answer:

o-na [289]3 years ago

6 0

Answer:

I may be wrong but I think 8 is your answer.

Step-by-step explanation:

(-1)^(3/7) x 128^(3/7)

-1 x 128^3/7

128^(3/7) = 8

= 8

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WHERE DO I PLACE THIS

allochka39001 [22]

The absolute value is the value of the number without the sign. So in this case it's 3 and -3

6 0

3 years ago

Read 2 more answers

La fórmula para calcular la distancia de "d" en pies, necesaria para detener un auto sobre una superficie con nieve es d=1/6 x^2

spin [16.1K]

D=1/6 x^2
150=1/6 x^2
*6 *6
900=x^2

8 0

3 years ago

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

7 0

3 years ago

4(9) - 2(7-3)^2<br><br><br>thanks if u help

MariettaO [177]

Answer:

Step-by-step explanation:

$36 -2(4)^{2}$

$36 - 2 \times 16$

$36 - 32$

$4$

7 0

3 years ago

The sum of 6 consecutive even numbers is 126

tatyana61 [14]

Start at 16 and count up by 2 until 26.

5 0

3 years ago