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andrew-mc [135]
3 years ago
10

If y ∝

smiddle" class="latex-formula"> and y = 4 when x = 2.5, find :
(a) y when x = 20

(b) x when y = 5
Mathematics
1 answer:
AlladinOne [14]3 years ago
6 0

Answer:

Step-by-step explanation:

Y = k1/x

4 = k 1/2.5

2.5 x 4 = k

10 = k

Now using this equation we will find the other answers

A) find y when x = 20

Y = k 1/x

Putting values

= 10(1/20) = 1/2 = 0.5

B) find x when y = 5

Y = k 1/x

Putting values

5 = 10 1/x

5x = 10

x = 10/5 = 2

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Tyler’s brother works in a shoe store. The store was selling a pair of shoes for $80. They now are on sale for $50. What is the
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3 years ago
PLEASE PLEASE HELP ME A pinecone drops from a tree branch that is 36 feet above the ground. The function h = -16t2 + 36 is used.
yKpoI14uk [10]

Answer:

T = 1.5s

Step-by-step explanation:

Hello,

To find the time the pincone hits the ground, we need to use the equation given.

Note that h = 0 when the pinecones hits the ground.

This question relates to motion under gravity.

h = -16t² + 36

0 = -16t² + 36

Make t² the subject of formula

16t² = 36

t² = 36 / 16

t² = 2.25

Take the square root of both sides

t = √(2.25)

t = 1.5s

The time it takes the pinecone to hit the ground is 1.5s.

3 0
3 years ago
Use the definition of continuity to determine whether f is continuous at a.
dmitriy555 [2]
f(x) will be continuous at x=a=7 if
(i) \displaystyle\lim_{x\to7}f(x) exists,
(ii) f(7) exists, and
(iii) \displaystyle\lim_{x\to7}f(x)=f(7).

The second condition is immediate, since f(7)=8918 has a finite value. The other two conditions can be established by proving that the limit of the function as x\to7 is indeed the value of f(7). That is, we must prove that for any \varepsilon>0, we can find \delta>0 such that

|x-7|

Now,


|f(x)-f(7)|=|5x^4-9x^3+x-8925|

Notice that when x=7, we have 5x^4-9x^3+x-8925=0. By the polynomial remainder theorem, we know that x-7 is then a factor of this polynomial. Indeed, we can write

|5x^4-9x^3+x-8925|=|(x-7)(5x^3+26x^2+182x+1275)|=|x-7||5x^3+26x^2+182x+1275|

This is the quantity that we do not want exceeding \varepsilon. Suppose we focus our attention on small values \delta. For instance, say we restrict \delta to be no larger than 1, i.e. \delta\le1. Under this condition, we have

|x-7|

Now, by the triangle inequality,


|5x^3+26x^2+182x+1275|\le|5x^3|+|26x^2|+|182x|+|1275|=5|x|^3+26|x|^2+182|x|+1275

If |x|, then this quantity is moreover bounded such that

|5x^3+26x^2+182x+1275|\le5\cdot8^3+26\cdot8^2+182\cdot8+1275=6955

To recap, fixing \delta\le1 would force |x|, which makes


|x-7||5x^3+26x^2+182x+1275|

and we want this quantity to be smaller than \varepsilon, so


6955|x-7|

which suggests that we could set \delta=\dfrac{\varepsilon}{6955}. But if \varepsilon is given such that the above inequality fails for \delta=\dfrac{\varepsilon}{6955}, then we can always fall back on \delta=1, for which we know the inequality will hold. Therefore, we should ultimately choose the smaller of the two, i.e. set \delta=\min\left\{1,\dfrac{\varepsilon}{6955}\right\}.

You would just need to formalize this proof to complete it, but you have all the groundwork laid out above. At any rate, you would end up proving the limit above, and ultimately establish that f(x) is indeed continuous at x=7.
5 0
3 years ago
What are the zeros of the function f(x) = x2 - 5x + 4?​
stich3 [128]

To find the zeros set the equation equal to zero:

x^2 - 5x +4 = 0

Factor:

(x-4) (x-1) = 0

Now solve each set of parenthesis so that they equal zero:

(x-4) = 0, x =4

(x-1) = 0, x = 1

The zeros are 1,4

5 0
3 years ago
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