The diameter of a circle is 10 in. Find its area in terms of pi

Mathematics

2 answers:

elena55 [62]2 years ago

4 0

Answer:

25 pi

Step-by-step explanation:

exis [7]2 years ago

4 0

78.54 is the area in terms of pi

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Two fewer than a number doubled is the same as the number decreased by 38. if n is "the number," which equation could be used to

Alona [7]

$\displaystyle \\ 
\text{We use the following equation:} \\ \\ 
 \frac{n}{2} = n - 38 \\ \\ 
We solve for n: \\ \\ 
 \frac{n}{2} = n - 38 ~~\Big| \times 2 \\ \\ 
n = 2 \times n - 2 \times 38 \\ \\ 
n = 2n - 76 \\ \\ 
2n - n = 76 \\ \\ 
\boxed{n = 76} \\ \\ 
\text{Verification: } \\ \\ 
\frac{n}{2} = n - 38 \\ \\ 
\frac{76}{2} = 76 - 38 \\ \\ 
\boxed{38 = 38 ~~~ Correct }$

8 0

3 years ago

Consider a fictitious company, Teslala, that produces a single type of electronic vehicle, Model Z. The demand for Model Z depen

grin007 [14]

The ratio X/Y for the fictitious company, Teslala, that produces a single type of electronic vehicle, Model Z is 10/pY.

<h3>What is a mathematical model?</h3>

A mathematical model is the model which is used to explain the any system, the effect of the components by study and estimate the functions of systems.

Consider a fictitious company, Teslala, that produces a single type of electronic vehicle, Model Z.

The demand for Model Z depends on the gasoline price (q) because customers tend to purchase an electronic vehicle as a substitute for vehicles that run on gasoline when the gasoline price increases.

The demand for Model Z is estimated as

$D(p) = 180 + 10q - 4p$

Here, <em>p</em> is the price of Model Z.

Consider the following two statements:

1. When the average gasoline price q increases by $1, the revenue-maximizing price p" increases by $X.

$D(p) = 180 + 10q - 4p\\D(p) = 180 + 10(q+1) - 4(p+X)$

2. When the average gasoline price q increases by $1, the demand at the revenue-maximizing price (i.e., D(p*)) increases by a factor of Y.

$D(p)+Y = 180 + 10q - 4p\\(D(p)+Y) = 180 + 10(q+1) - 4(p+X)\\Y= 180 + 10(q+1) - 4(p+X)-D(p)\\$

Put the value of demand at the revenue-maximizing price as,

$Y= 180 + 10(q+1) - 4(p+X)-(180 + 10q - 4p)\\Y= 180 + 10q+10 - 4p-pX-180 - 10q + 4p\\Y= 10 -pX\\1=\dfrac{10}{Y}-\dfrac{pX}{Y}\\\dfrac{10}{Y}=\dfrac{pX}{Y}\\\dfrac{10}{pY}=\dfrac{X}{Y}$