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Damm [24]
4 years ago
10

The expression 20l + 25g - 10 gives the number of dollars that Josie makes from mowing lawns and raking gardens.

Mathematics
1 answer:
vodka [1.7K]4 years ago
4 0
20l + 25g -10
replace G with 4 and L with 3

20(3) + 25(4) - 10

60 + 100 - 10

160 - 10

150

he made $150
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The diameter of the base of the cone measures 8 units.
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Answer:   32\pi cubic units.

Step-by-step explanation:

You can use this formula for calculate the volume of a cone:

V_{cone}=\frac{1}{3}\pi r^2h

Where "r" is the radius and "h" is the height.

You know that the diameter of the base of the cone measures 8 units, then, the radius can be found by dividing the diameter by 2:

r=\frac{8units}{2}\\\\r=4units

Since you already know that height and  the radius, you can substitute them into the formula. Then, the volume of this cone is:

V_{cone}=\frac{1}{3}\pi (4units)^2(6units)

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Sidana [21]

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Step-by-step explanation:

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3 years ago
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An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
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3 years ago
Sanjay read 56 pages of his book this weekend.This is 35% of the pages in the book.How many pages are in Sanjays book?
Maurinko [17]
1. 56 : 35 = 1.6
2. 1.6* 100 = 160
3. The answer is 160
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4 years ago
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