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max2010maxim [7]
3 years ago
13

Which similarity statements are true? Check all that apply. △JKL ~ △KML △JKM ~ △JKL △JKM ~ △KML △JMK ~ △JKL △JMK ~ △KML

Mathematics
2 answers:
Aliun [14]3 years ago
5 0
The answer:

according to the image, the main theorem concerning right triangle similarity is as follow:
the altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to the original triangle and each other, 

in our case, KL is the altitude, and by applying theorem, we get three triangles that are similar:
therefore:
<span>△JKL ~ △JKM
</span><span>△JKM ~ △JKM
</span><span>△JMK ~ △KML</span>
loris [4]3 years ago
3 0
The answeres are 1) JKL~ KML ,4) JMK~JKL,and 5) JMK~KML 
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ehidna [41]

Answer:

Step-by-step explanation:

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3 years ago
Mark sold x pens for $2 each. He received less than $80 from the sale. Find the greatest possible number of pens he sold
Crank
$80-$2= $78 is the number less than $80, since the pens cost $2

78/2= 39 pens is the greatest number of pens sold

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3 years ago
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Jake has proved that a function, f(x), is a geometric sequence. How did he prove that?
Mashutka [201]

Answer:

He showed that f(n) ÷ f(n - 1) was a constant ratio.

Given that Jake has proved that a function f(x) is a geometric sequence.

GEOMETRIC SEQUENCE:  A geometric sequence is a sequence of numbers where each term is found by multiplying the preceding term by a constant called the common ratio, r.

So, in Jame's proof, he showed that each term is multiplied by a constant to get the next term.

That is, if 'c' is the constant that was used in the proof, then we must have

This implies that

Therefore, he showed that  f(n) ÷ f(n - 1) was a constant ratio.

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3 years ago
Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th
trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

    t = 20: 999 fishes

    t = 30: 1168 fishes

(c)

P(\infty) = 1200

Step-by-step explanation:

Given

P(t) =\frac{d}{1+ke^-{ct}}

d = 1200\\k = 11\\c=0.2

Solving (a): Fishes at t = 0

This gives:

P(0) =\frac{1200}{1+11*e^-{0.2*0}}

P(0) =\frac{1200}{1+11*e^-{0}}

P(0) =\frac{1200}{1+11*1}

P(0) =\frac{1200}{1+11}

P(0) =\frac{1200}{12}

P(0) = 100

Solving (a): Fishes at t = 10, 20, 30

t = 10

P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483

t = 20

P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999

t = 30

P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168

Solving (c): \lim_{t \to \infty} P(t)

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of e^{-ct} decreases

This implies that:

{t \to \infty} = {e^{-ct} \to 0}

So:

The value of P(t) for large values is:

P(\infty) = \frac{1200}{1 + 11 * 0}

P(\infty) = \frac{1200}{1 + 0}

P(\infty) = \frac{1200}{1}

P(\infty) = 1200

5 0
2 years ago
2350 million in standard form
topjm [15]

Answer:

in this problem we have

2,350 million

Remember that

1 million=1,000,000

so

2,350 million=2,350*1,000,000=2,350,000,000

convert to standard form

2,350,000,000=2.35*10^{9}

therefore

the answer is

2.35*10^{9}

Step-by-step explanation:

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