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2 years ago

X2 +4+5y3+2x2-2x2-6+1

Mathematics

2 answers:

Andreyy892 years ago

6 0

Answer:

$x^{3} +5y^{3} -1$

Step-by-step explanation:

Verdich [7]2 years ago

6 0

Answer: x^2-1+5y^3

Step-by-step explanation:

x^2+4+5y^3-6+1 (2x^2-2x^2 cancel each other.

x^2-1+5y^3 (4-6+1= -1)

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X2 + 5x - 50 A) (x + 25)(x - 2) B) (x - 5)(x + 10) C) (x - 10)(x + 5) D) (x + 2)(x - 25)

olga2289 [7]

Answer:

The correct answer is B) (x - 5)(x + 10)

Step-by-step explanation:

The constants at the end of the parenthesis need to add to the middle term

10 + -5 = 5

And they need to multiply to the end term

-5 * 10 = -50

This makes it the correct choice.

3 0

2 years ago

If the ratio of side lengths of similar polygons is 6:11, what is the ratio of perimeters

algol13

Answer:

6 : 11

Step-by-step explanation:

the ratio 6 : 11 applies to all linear measure in the similar polygons

Both side lengths and perimeter are linear, hence

ratio of both is 6 : 11

8 0

2 years ago

Intr un album numarul de fotografii s a micsorat cu 50%. Cu cat trebuie sa se mareasca pentru a reveni la numarul initial? Raspu

hodyreva [135]

Răspuns:

100%

Explicație pas cu pas:

Permiteți numărul inițial de fotografii = x

x - 50% din x = x - 0,5x = 0,5x

0,5x + câte% din 0,5x = x

x - 0,5x = y% de 0,5x

0,5x / 0,5x = 1

Prin urmare,

100% = 1

Prin urmare, x trebuie mărit cu 100% pentru a reveni la numărul original.

3 0

2 years ago

What's 5x3833 in an equation divided by two

Xelga [282]

You mean the answer of 5x3833 divided by two?

7 0

2 years ago

A bacteria colony increases in size at a rate of 4.0581e1.6t bacteria per hour. If the initial population is 36 bacteria, find t

Harrizon [31]

Answer:

1560

Step-by-step explanation:

The rate of Increase of the population (P) of the bacteria is given as:

$\frac{dP}{dt} =4.058e^{1.6t}$

$dP =4.058e^{1.6t}}dt\\Taking\: Integrals\\\int dP =4.058 \int e^{1.6t}dt\\P(t)=\frac{4.058}{1.6} (e^{1.6t} +K)\\P(t)=2.53625 (e^{1.6t} +K)$

Where k is a constant of Integration.

At t=0, P(t)=36

$36=2.53625 (e^{1.6*0} +K)\\36=2.53625 (e^{0} +K)\\36=2.53625 (1 +K)\\36=2.53625 +2.53625K\\K=13.19$

Therefore:

$P(t)=2.53625 (e^{1.6t} +13.19)\\At \: t=4\\P(4)=2.53625 (e^{1.6*4} +13.19)\\=1559.88\\P(4)=1560$

3 0

3 years ago

X2 +4+5y3+2x2-2x2-6+1​

X2 +4+5y3+2x2-2x2-6+1