Can someone help me with this question

Mathematics

1 answer:

stira [4]3 years ago

8 0

The answer is 90 degrees.

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What is the greatest common factor of the terms in the polynomial 4x^4-32x^3-60x^2?

jonny [76]

Answer:

$\mathrm{Factor}\:4x^4-32x^3-60x^2:\quad 4x^2\left(x^2-8x-15\right)\\$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c\\x^3=xx^2\\x^4=x^2x^2\\=4x^2x^2-32xx^2-60x^2\\\mathrm{Rewrite\:}60\mathrm{\:as\:}4\cdot \:15\\\mathrm{Rewrite\:}32\mathrm{\:as\:}4\cdot \:8\\=4x^2x^2-4\cdot \:8xx^2-4\cdot \:15x^2\\\mathrm{Factor\:out\:common\:term\:}4x^2\\=4x^2\left(x^2-8x-15\right)$

<em>Hope this helps and have a great day!!!</em>

<em> $Sofia$ </em>

4 0

3 years ago

Find the common ratio for this geometric sequence.

Nonamiya [84]

Answer:

i believe it is B.

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3 years ago

If N is the population of the colony and t is the time in days, express N as a function of t. Consider Upper N 0 is the origina

lesya [120]

Answer:

(a) $N(t)=Noe^{kt}$

(b)5,832 Mosquitoes

Step-by-step explanation:

(a)Given an original amount $N_o$ at t=0. The population of the colony with a growth rate $k \neq 0$ , where k is a constant is given as:

$N(t)=Noe^{kt}$

(b)If $N_o=1000$ and the population after 1 day, N(1)=1800

Then, from our model:

N(1)=1800

$1800=1000e^{k}\\$Divide both sides by 1000\\e^{k}=1.8\\$Take the natural logarithm of both sides\\k=ln(1.8)$

Therefore, our model is:

$N(t)=1000e^{t*ln(1.8)}\\N(t)=1000\cdot1.8^t$

In 3 days time

$N(3)=1000\cdot1.8^3=5832$

The population of mosquitoes in 3 days time will be approximately 5832.

(c)If the population N(t)=20,000,we want to determine how many days it takes to attain that value.

From our model

$N(t)=1000\cdot1.8^t\\20000=1000\cdot1.8^t\\$Divide both sides by 1000\\20=1.8^t\\$Convert to logarithm form\\Log_{1.8}20=t\\\frac{Log 20}{Log 1.8}=t\\ t=5.097\approx 5\; days$