There are currently 12 frogs in a large) pond. The frog population grows exponentially, tripling every 8 days.

Help please !!!!!!!!!

Anni [7]

I’m not really sure about the answer but I’m doing the same think n this should help sorry ‍♀️

7 0

2 years ago

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Suppose that f(t) is continuous and twice-differentiable for t≥0. Further suppose f″(t)≥9 for all t≥0 and f(0)=f′(0)=0. Using th

Paha777 [63]

Give me a heart and I'll tell you the answer

5 0

2 years ago

For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perim

Marina86 [1]

Area of the shaded region $$=36(\pi -2)$ square cm

Perimeter of the shaded region $=6 (\pi + 2\sqrt 2)$ cm

Solution:

Radius of the quarter of circle = 12 cm

Area of the shaded region = Area of quarter of circle – Area of the triangle

$$=\frac{1}{4} \pi r^2 - \frac{1}{2} bh$

$$=\frac{1}{4} \pi \times 12^2 - \frac{1}{2} \times 12 \times 12$

$$=36\pi -72$

$$=36(\pi -2)$ square cm.

Area of the shaded region $$=36(\pi -2)$ square cm

Using Pythagoras theorem,

$AC^2=AB^2+BC^2$

$AC^2=12^2+12^2$

$AC^2=288$

Taking square root on both sides of the equation, we get

$AC= 12\sqrt 2$ cm

Perimeter of the quadrant of a circle = $\frac{1}{4} \times 2\pi r$

$$=\frac{1}{4} \times 2 \times \pi \times 12$

$$=6 \pi$ cm

Perimeter of the shaded region = $6 \pi + 12\sqrt 2$ cm

$=6 (\pi + 2\sqrt 2)$ cm

Hence area of the shaded region $$=36(\pi -2)$ square cm

Perimeter of the shaded region $=6 (\pi + 2\sqrt 2)$ cm

6 0

2 years ago

Tony's Mom gives him a weekly allowance. Due to his bad behavior, his allowance has been shrinking at a rate of 2 3/10 dollars e

nikitadnepr [17]

Answer:

-3 2/7 per day

Step-by-step explanation:

-2 3/10 then put that all over 3 1/2

-23/10 divided by 7/2

-23/10 *21/7 = -23/7= -3 2/7 per day

the 10 cross out and make a 5

i should be right

7 0

2 years ago

Read 2 more answers

HELP PLEASE!!!!!!!!!!!

emmainna [20.7K]

Answer:

Part 1) $y=-x^{2}$ ---> Translated up by 1 units

Part 2) $y=x^{2}+1$ ---> Reflected across the x-axis

Part 3) $y=-(x+1)^{2}-1$ ---> Translated left by 1 unit

Part 4) $y=-(x-1)^{2}-1$ ----> Translated right by 1 unit

Part 5) $y=-x^{2}-1$ ----> Reflected across the y-axis

Part 6) $y=-x^{2}-2$ ----> Translated down by 1 unit

Step-by-step explanation:

we know that

The parent function is

$y=-x^{2}-1$ ----> this is a vertical parabola open downward with vertex at (0,-1)

<em>Calculate each case</em>

Part 1) Translated up by 1 unit

The rule of the translation is

(x,y) -----> (x,y+1)

so

(0,-1) ----> (0,-1+1)

(0,1) ----> (0,0) ----> the new vertex

The new function is equal to

$y=-x^{2}$

Part 2) Reflected across the x-axis

The rule of the reflection is

(x,y) -----> (x,-y)

so

(0,-1) ----> (0,1) ----> the new vertex

The new function is equal to

$y=x^{2}+1$

Part 3) Translated left by 1 unit

The rule of the translation is

(x,y) -----> (x-1,y)

so

(0,-1) ----> (0-1,-1)

(0,1) ----> (-1,-1) ----> the new vertex

The new function is equal to

$y=-(x+1)^{2}-1$

Part 4) Translated right by 1 unit

The rule of the translation is

(x,y) -----> (x+1,y)

so

(0,-1) ----> (0+1,-1)

(0,1) ----> (1,-1) ----> the new vertex

The new function is equal to

$y=-(x-1)^{2}-1$

Part 5) Reflected across the y-axis

The rule of the reflection is

(x,y) -----> (-x,y)

so

(0,-1) ----> (0,-1) ----> the new vertex

The new function is equal to

$y=-x^{2}-1$

Part 6) Translated down by 1 unit

The rule of the translation is

(x,y) -----> (x,y-1)

so

(0,-1) ----> (0,-1-1)

(0,1) ----> (0,-2) ----> the new vertex

The new function is equal to

$y=-x^{2}-2$

3 0

2 years ago