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Nadusha1986 [10]
3 years ago
6

30 POINTS!!! A catfish weights about 0.2 pound and its weight increases by about 15% each week. Write an exponential equation th

at represents the weight of the catfish after t weeks.
Mathematics
1 answer:
dezoksy [38]3 years ago
7 0

Answer:

0.30 15 times 2 easy 30 and this percent so 0.30

Step-by-step explanation:

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Which pair of statements about the lengths of two sides of the triangle is true?
Georgia [21]

Answer:

B

Step-by-step explanation:

7 0
3 years ago
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solve the equation: the booster club of the baseball team bought pizza to sell at a concession stand. the school donated three p
labwork [276]

Answer:

Booster club purchased 5 pies.

Step-by-step explanation:

Let the number of pies purchased by booster club be represented by n. The school also donated 3 pies to sell.

So that,

Total number of pies = n + 3

Given that there was 64 pieces of pies to sell after cutting each pie into 8 pieces.

Thus,

8( n + 3) = 64

8n + 24 = 64

8n = 64 - 24

8n = 40

Divide both sides by 8 to have,

n = 5

The number of pies purchased by booster club is 5.

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3 years ago
Parallel lines in space are coplanar.<br> a. always<br> b. sometimes<br> c. never
Katyanochek1 [597]
B. Sometimes because in learned that it isn't always that way.
3 0
3 years ago
Now Ann is x years old and her sister is eight years younger than her. After five years , Anne's age is twice the age of her sis
svp [43]

Answer:

You can refer to the given attachment

7 0
2 years ago
First you to find the worksheet and download it<br> plase I need help
Goshia [24]

Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = 3^{x + 2}

The asymptote is given as x → -∞, f(x) = 3^{x + 2} → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = 3^{0 + 2} = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = 5^{1  - x}

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = 5^{1  - 0} = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = 2^{x + 3} - 1

The asymptote is 2^{x + 3}  → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) =  2^{0 + 3} - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, 2^{x + 3} - 1 = 0

2^{x + 3} = 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f) f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2

The asymptote is \left (\dfrac{1}{2} \right)^{x - 2} → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6

The y-intercept is (0, 6)

7 0
2 years ago
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