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snow_lady [41]
2 years ago
10

(2) Answer.. A frog is sitting at the bottom of a well 31 m deep. During the day when he climbs the wall but is only able to cli

mb 6 m but at night with the heavy rains he slides back down 4 m. Identify how long it will take the frog to get out of the well if the same pattern continues everyday?​
Mathematics
1 answer:
valentina_108 [34]2 years ago
8 0

Answer:

Step-by-step explanation:

1 day= 6-4= 2 meters

31/2 meters= 15.5 days

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(a) Each student in a pottery class is responsible for cleaning the area around his or her workspace at the end of the lesson. E
ArbitrLikvidat [17]
Ok this is not as hard as it seems

(4.5 feet)^2 times the number of students (13) 

20.25 x 13

263.25 is the correct answer



I hope it helps :<
5 0
3 years ago
Using the formula in model 1, choose the correct answers for the new balance and amount of interest earned in the following comp
melisa1 [442]
To find the total amount
The formula is
A=p (1+r)^t
A total amount?
P present value 1050
R interest rate 0.06
T time 25 years
A=1,050×(1+0.06)^(25)=4,506.46

Interest amount
I=A-p
I=4,506.46−1,050
I=3,456.46
5 0
3 years ago
What is the area of triangle DEF?
FrozenT [24]

Answer:

A=2\sqrt{14}\ units^2

Step-by-step explanation:

we know that

Heron's Formula is a method for calculating the area of a triangle when you know the lengths of all three sides.  

so

A=\sqrt{s(s-a)(s-b)(s-c)}

where

a, b and c are the length sides of triangle

s is the semi-perimeter of triangle

we have

a=5\ units,b=6\ units,c=3\ units

<em>Find the semi-perimeter s </em>

s=\frac{5+6+3}{2}=7\ units

Find the area of triangle

A=\sqrt{7(7-5)(7-6)(7-3)}

A=\sqrt{7(2)(1)(4)}

A=\sqrt{56}\ units^2

simplify

A=2\sqrt{14}\ units^2

3 0
2 years ago
Find the exact length of the curve. 36y2 = (x2 − 4)3, 5 ≤ x ≤ 9, y ≥ 0
IrinaK [193]
We are looking for the length of a curve, also known as the arc length. Before we get to the formula for arc length, it would help if we re-wrote the equation in y = form.

We are given: 36 y^{2} =( x^{2} -4)^3
We divide by 36 and take the root of both sides to obtain: y = \sqrt{ \frac{( x^{2} -4)^3}{36} }

Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain: y =  \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}

Let's leave that for the moment and look at the formula for arc length. The formula is L= \int\limits^c_d {ds} where ds is defined differently for equations in rectangular form (which is what we have), polar form or parametric form.

Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows: ds= \sqrt{1+( \frac{dy}{dx})^2 } dx

As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.

As you can see from the formula we need to find dy/dx and square it. Let's do that now.

We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here x^2-4). More formally, we can let u=x^{2} -4 and then consider the derivative of u^{3/2}du. Either way, we obtain,

\frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2}

Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4}

This means that in our case:
ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{x^4-4 x^{2}+4 }{4}} dx
ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx
ds=  \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx

Recall, the formula for arc length: L= \int\limits^c_d {ds}
Here, the limits of integration are given by 5 and 9 from the initial problem (the values of x over which we are computing the length of the curve). Putting it all together we have:

L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x evaluated from 9 to 5 (I cannot seem to get the notation here but usually it is a straight line with the 9 up top and the 5 on the bottom -- just like the integral with the 9 and 5 but a straight line instead). This means we plug 9 into the expression and from that subtract what we get when we plug 5 into the expression.

That is, [(\frac{1}{2}) ( \frac{9^3}{3}) -9]-([(\frac{1}{2}) ( \frac{5^3}{3}) -5]=( \frac{9^3}{6}-9)-( \frac{5^3}{6}-5})=\frac{290}{3}


8 0
3 years ago
I need help on this question
Flura [38]

Answer:

you are adding a dollar and 50 cents each time

Step-by-step explanation:

then find your y and x and plot it

6 0
3 years ago
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