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aleksandrvk [35]

2 years ago

8

The area, a, of an ellipse can be determined using the formula a = πxy, where x and y are half the lengths of the largest and sm

allest diameters of the ellipse.
Which is an equivalent equation solved for y?

y = a · πx
y = a + (πx)
y = a – πx
y = a ÷ (πx)

2 answers:

Mrrafil [7]2 years ago

7 0

It is the 3 rd if I’m wrong I’m sorry

Tems11 [23]2 years ago

7 0

The answer to your question would be <u>B</u> hope this helps pls mark brainliest.

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Find the number of liters in 12.8 gal of gasoline. Round to the nearest hundredth

enot [183]

Answer:

48.45327

Step-by-step explanation:

12.8 x 3.785 = 48.454327

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2 years ago

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Wynn needs to find the center of the data set shown on the dot plot. The dot plot has dots. The dot plot has dots to the left of

Arisa [49]

Answer:

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2 years ago

What additional information could be used to prove ΔABC ≅ ΔMQR using SAS? Select two options. m∠A = 64° and AB = MQ = 31 cm CB =

enyata [817]

Answer:

m∠R = 60° and AB ≅ MQ

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Step-by-step explanation:

Given data :

Prove ΔABC ≅ ΔMQR using SAS

The missing information to prove ΔABC ≅ ΔMQR using SAS

m∠R = 60° and AB ≅ MQ

m∠Q = 56° and CB ≅ RQ

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2 years ago

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Pleaseeee answer quick WILLL GIVE BRAINLIEST!!!!!! Given the expression: 5x10 − 80x2 Part A: Rewrite the expression by factoring

sladkih [1.3K]

Answer:

$5\,x^{10}-80\,x^2=5\,x^2\,(x^4+4)(x^2-2)(x^2+2)$

Step-by-step explanation:

First, extract the greatest common factor of the numerical as well as the variable parts:

the GCF of 5 and 80 is "5", and the GCF between $x^{10}$ and $x^2$ is $x^2$ , then:

$5\,x^{10}-80\,x^2=5\,x^2\,(x^8-16)$

Now, we can re-write the binomial in parenthesis as a difference o squares to be able to factor out even further:

$x^8-16=(x^4)^2-4^2=(x^4-4)(x^4+4)$

we now repeat the process for the binomial difference:

$(x^4-4)=((x^2)^2-2^2)=(x^2-2)(x^2+2)$

So finally the complete factorization is:

$5\,x^{10}-80\,x^2=5\,x^2\,(x^4+4)(x^2-2)(x^2+2)$

4 0

2 years ago

1. Consider the population model dP dt = 0.2P 1 − P 135 , where P(t) is the population at time t. (a) For what values of P is th

Kryger [21]

Answer:

a

$P = 0$ OR $P = 135$

b

$P > 0$ and $P < 135$

OR

$P > 0$ and $P < 135$

c

Generally the carrying capacity is can be defined as the highest amount of population and environment can support for an unlimited duration or time period

d

$P = 67.5$

Step-by-step explanation:

From the question we are told that

The population model is $\frac{dP}{dt} = 0.2P(1 - \frac{P}{135} )$

Generally at equilibrium

$\frac{dP}{dt} = 0$

So

$0.2P = 0$

=> $P = 0$

Or

$(1 - \frac{P}{135} ) = 0$

=> $P = 135$

Thus at equilibrium P = 0 or P = 135

Generally when the population is increasing we have that

$\frac{dP}{dt} > 0$

So

$0.2P > 0$

=> $P > 0$

and

$(1 - \frac{P}{135} ) > 0$

$P < 135$

Now when the first value of P i.e $P< 0$ for $\frac{dP}{dt} > 0$

$P_2 > 135$

So when population increasing the values of P are

$P > 0$ and $P < 135$

OR

$P > 0$ and $P < 135$

So to obtain initial values of P where the population converge to the carrying capacity as $t \to [\infty]$

The rate equation can be represented as

$\frac{dP}{dt} = \frac{1}{5}P (1 - \frac{P}{135} )$

So we will differentiate the equation again we have that

$\frac{d^2 P}{dt^2} = \frac{(1 - \frac{P}{135} )}{5} - \frac{P}{675}$

Now as $t \to [\infty]$

$\frac{d^2 P}{dt^2} \to 0$

So

$\frac{(1 - \frac{P}{135} )}{5} - \frac{P}{675} = 0$

=> $\frac{(1 - \frac{P}{135} )}{5} = \frac{P}{675}$

=> $P = 67.5$

5 0

2 years ago