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3 years ago

Find the product. (x2+6x +9)(x?) - What’s the product

Mathematics

2 answers:

Annette [7]3 years ago

7 0

(x²+6x+9)x=x(x+3)²

.............

lesya692 [45]3 years ago

4 0

Answer:

Step-by-step explanation:

(x² + 6x + 9)(x)= x²*x + 6x*x + 9*x = x³ + 6x² + 9x

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'Nine less than the product of four and a number

maxonik [38]

Answer:

(4 x X)-9

Step-by-step explanation:

I think this is correct pls tell me if this is wrong

Hope this helps

7 0

3 years ago

At sea level, earths atmosphere exerts a pressure of 1 atmosphere. Atmospheric pressure P (in atmospheres) decreases with altitu

PtichkaEL [24]

Answer:

Initial Amount=1

Decay Factor=0.99987

Decay Rate=0.013%

Step-by-step explanation:

If P = 0.99987ᵃ

I. Initial amount

The initial amount is when a=0

P = 0.99987⁰ = 1

The Pressure at Sea Level=1

II. Decay factor

This is like the decay constant. From the model, P = 0.99987ᵃ, the decay constant/factor is 0.99987.

III. Decay Rate = 1-0.99987 = 0.00013

Expressed in Percentage = 0.013%

5 0

3 years ago

Read 2 more answers

Find the following integral

ololo11 [35]

There's nothing preventing us from computing one integral at a time:

$\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2$

$\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2$

$\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx$

Expand the integrand completely:

$x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x$

Then

$\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}$