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neonofarm [45]
3 years ago
8

There are 20 alligators in the swamp. Each year, the number of alligators increases by 25%. There are 25 crocodiles in the swamp

. Each year, 10 new crocodiles join the swamp.
Part A: Write functions to represent the number of alligators and crocodiles in the swamp throughout the years. (4 points)

Part B: How many alligators are in the swamp after 4 years? How many crocodiles are in the swamp after the same number of years? (2 points)

Part C: After approximately how many years is the number of alligators and crocodiles the same? Justify your answer mathematically. (4 points)
Mathematics
1 answer:
myrzilka [38]3 years ago
7 0
Part A: f(x)=(5)x+20(alligators) and f(x)=(10)x+25(crocodiles)

Part B: f(x)=(5)(4)+20(alligators) and f(x)=(10)(4)+25(crocodiles)
Alligators: 40 and Crocodiles: 65

Part C: None because the rate of the alligators won't catch up with the rate of crocodiles.
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Write the percent as a fraction and as a decimal. 3/5% I'm really stumped.
Karo-lina-s [1.5K]

3/5 = 0.6 =60% i hope it helps.

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3 years ago
Can any one tell me this i have to do this for my tomorrow school
Rashid [163]

Answer:

Step-by-step explanation:

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7 0
3 years ago
the height of a triangle is 7cm greater than its base and its area is 15cm squared. find the value of its height
Rudiy27

Answer:

Step-by-step explanation:

Let base be b, height be h

h = 7+b or we can say that b = h -7

A = 15 cm²

General formula for area of a triangle

A = b · h

Substitute what we know to find the height h

15 = (h-7) · h

15 = h² -7h

0 = h²-7h -15

h = (7 ±√49 + 4·1·15)/2

h = (7 ±√109)/2

h ≈ -1.72 reject this solution because height cannot be negative

h ≈ 8.72 cm

5 0
2 years ago
Can you prove it??<br>it's hard,I tried but couldn't solve this.
BARSIC [14]

I'll abbreviate s=\sin\theta and c=\cos\theta, so the identity to prove is

\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)

On the left side, we can simplify a bit:

\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}

\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}

Then

\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)

So the establish the original equality, we need to show that

\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s

Combine the fractions:

\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{2-2s}{c^2-s^2+2s-1}

We can rewrite the denominator as

c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1

then using the fact that c^2+s^2=\cos^2\theta+\sin^2\theta=1, we get

1-2s^2+2s-1=2s-2s^2

so that we have

\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s

as desired.

6 0
4 years ago
Alex brought a car for $9000. The sales tax rate on the car is 3 %. How much sales tax did Alex pay
Nookie1986 [14]
9,000 * 3% = 270
So it would be $270 in sales tax
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