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kari74 [83]
3 years ago
14

Solve for x: −2(x − 2)2 + 5 = 0

Mathematics
2 answers:
Dimas [21]3 years ago
6 0

Answer: x = 3.58, 0.42

Step-by-step explanation: i took the test lol

Licemer1 [7]3 years ago
5 0

Option A

The values of x are 3.58 or 0.42

<u>Solution:</u>

Given, equation is -2(x-2)^{2}+5=0

We have to solve for the x and choose from the options given,

Now, take the equation,

\begin{array}{l}{\rightarrow-2(x-2)^{2}+5=0} \\\\ {\rightarrow-2(x-2)^{2}=-5}\end{array}

Negative sign cancels on both sides

\begin{array}{l}{\rightarrow 2(x-2)^{2}=5} \\\\ {\rightarrow(x-2)^{2}=\frac{5}{2}}\end{array}

Taking square root on both sides, we get

\begin{array}{l}{\rightarrow x-2=\sqrt{\frac{5}{2}}} \\\\ {\rightarrow x-2=\pm 1.58} \\\\ {\rightarrow x=2 \pm 1.58}\end{array}

\begin{array}{l}{\rightarrow x=2+1.58 \text { or } 2-1.58} \\\\ {\rightarrow x=3.58 \text { or } 0.42}\end{array}

Hence, the values of x are 3.58 or 0.42, so option A is correct.

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navik [9.2K]

The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.

Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).

(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get

P(t)=40

(50(1+0.05t))/(2+0.01t)=40

Cross multiply both sides, we get

50(1+0.05t)=40(2+0.01t)

Apply the distributive property a(b+c)=ab+ac, we get

50+2.5t=80+0.4t

Subtract 0.4t and 50 from both sides, we get

50+2.5t-0.4t-50=80+0.4t-0.4t-50

2.1t=30

Divide both sides with 2.1, we get

t=14.29 months

(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get

\begin{aligned}\lim_{t \rightarrow \infty}P(t)&=\lim_{t \rightarrow \infty}\frac{50(1+0.05t)}{2+0.01t}\\ &=\lim_{t \rightarrow \infty}\frac{50(\frac{1}{t}+0.05)}{\frac{2}{t}+0.01}\\ &=50\times \frac{0.05}{0.01}\\ &=250\end

(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.

Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.

Learn more about limiting factor from here brainly.com/question/18415071.

#SPJ1

8 0
2 years ago
A company sells fruit cups in packs of 4. The packs currently weigh 20 oz. The company plans to reduce the weight of each cup by
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Answer:

4 ( 5 - n )

Step-by-step explanation:

The expression:

20 – 4n

Task:  Rewrite the expression for the new weight as a product of two factors.

To do this we simply have to find the highest common factor (HCF) between 20 and 4n

20 = 4 * 5

4n = 4 * n

The HCF = 4

Expressing as a product of two factors;

4 ( 5 - n )

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2 years ago
What is the product of the polynomials below?<br> (3x2 - 2x - 3)(5x2 + 4x + 5)
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Answer

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Explain what SOHCAHTOA means.
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Answer:

It's a mnemonic device to help you remember the three basic trig ratios used to solve for missing sides and angles in a right triangle. It's defined as: SOH: Sin(θ) = Opposite / Hypotenuse. CAH: Cos(θ) = Adjacent / Hypotenuse.

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The ratio of the number of dogs to the number of cats is 4:5. There are 270 animals on the rescue farm. The number of dogs is 4/
ale4655 [162]

Answer:

True

Step-by-step explanation:

In this question, we are expected to evaluate the validity of a particular proposition.

To evaluate this validity, what we do is that we check if the calculations as projected on the question are correct.

Let’s proceed!

Number of dogs to cat ratio = 4:5

Number of animals is 270;

Number of dogs will be 4/9 * 270 = 120

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Now let’s check if the number of dogs is 4/5 that of cats.

Obviously 4/5 * 150 = 120

This makes true, the statement

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