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3 years ago

8

Qual o resultado da subtração dos dois polinômios abaixo? (2x2 + 4x) (–2x2 + x – 6)

1 answer:

Len [333]3 years ago

7 0

Answer:

-4x^4-6x^3-8x^2-24x

Step-by-step explanation:

$\left(2x^2+4x\right)\left(-2x^2+x-6\right)\\$

distribuir parênteses

$=2x^2\left(-2x^2\right)+2x^2x+2x^2\left(-6\right)+4x\left(-2x^2\right)+4xx+4x\left(-6\right)$

Aplicar regras de menos mais ; $+\left(-a\right)=-a$

$=-2\times\:2x^2x^2+2x^2x-2\times\:6x^2-4\times\:2x^2x+4xx-4\times\:6x$

simplificar

$-2\times\:2x^2x^2+2x^2x-2\times\:6x^2-4\times\:2x^2x+4xx-4\times\:6x:\\\quad -4x^4-6x^3-8x^2-24x$

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Plz help me with this question

irga5000 [103]

Answer:

122

Step-by-step explanation:

$x^6+y^2$

Plug in 1 for x and 11 for y

$=1^6+11^2$

1 to the power of any number will always equal 1

$=1+11^2\\=1+121\\=122$

I hope this helps!

7 0

3 years ago

1) Determine if the side lengths given form a triangle

Margaret [11]

Using the triangle inequality theorem, we can figure this out.

2- Yes

3- No

4- No

5- Yes

6- Yes

6 0

3 years ago

A triangle has side lengths of 13 in, 29 in, and 30 in. Classify it as acute, obtuse, or right.

kenny6666 [7]

Acute

Angle ∠ A = α = 25.385° = 25°23'6″ = 0.443 rad

Angle ∠ B = β = 73.004° = 73°14″ = 1.274 rad

Angle ∠ C = γ = 81.611° = 81°36'41″ = 1.424 rad

8 0

3 years ago

AC if TC = 20q + 10q^2?

Alexus [3.1K]

Answer:

AC = (20+ 10q)

Step-by-step explanation:

Given that,

Total cost, TC = 20q + 10q²

We need to find AC i.e. average cost.

It can be solved as follows :

$AC=\dfrac{TC}{q}\\\\AC=\dfrac{20q + 10q^2}{q}\\\\AC=\dfrac{q(20+ 10q)}{q}\\\\AC={(20+ 10q)}$

So, the value of AC is (20+ 10q).

4 0

3 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago