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2 years ago

Learn with an example

Mathematics

1 answer:

mafiozo [28]2 years ago

7 0

Answer:

-8

Step-by-step explanation:

In order to find the answer to this question you need to use the slope formula, substitute, subtract, and simplify if needed.

$(9,5),(4,1) = (x^1,y^1), (x^2,y^2)$

$m = \frac{y^2-y^1}{x^2-x^1} = \frac{9-1}{4-5}$

$9-1=8$

$4-5=-1$

$= -8$

Hope this helps.

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There are nickels, dimes, quarters in a large piggy bank, for every 2 nickels there are 3 dimes, for every 2 dimes there are 5 q

Alex

There are 80 nickels and 120 dimes and 300 quarters in piggy bank

Solution:

Given that,

For every 2 nickels there are 3 dimes ,

So, ratio of nickel to dime will be 2 : 3

Also for every 2 dimes there are 5 quarters

So, ratio of dime to quarters is 2 : 5

Let us first find the ratio of nickles to dimes to quarters

nickel : dime = 2 : 3

dime : quarters = 2 : 5

$\frac{nickel}{dime} = \frac{2}{3}$

$nickel = \frac{2}{3} dime$

$\frac{dime}{quarter} = \frac{2}{5}$

$quarter = \frac{5}{2}dime$

So,

$nickel : dime : quarter = \frac{2}{3}dime : dime : \frac{5}{2}dime\\\\nickel : dime : quarter = \frac{2}{3} : 1 : \frac{5}{2}\\\\nickel : dime : quarter = 4 : 6 : 15$

Thus nickel : dime : quarters = 4 : 6 : 15

Now, let the number of nickel be 4x

Let the number of dimes be 6x

Let the number of quarters be 15x

From given question,

There are 500 coins in all.

number of nickel + number of dimes + number of quarters = 500

4x + 6x + 15x = 500

25x = 500

x = 20

Thus,

number of nickel = 4x = 4(20) = 80

number of dimes = 6x = 6(20) = 120

number of quarters = 15x = 15(20) = 300

Thus there are 80 nickels and 120 dimes and 300 quarters in piggy bank

3 0

3 years ago

P and Q vary inversely. If P is 16.4 "when Q is 123, find an equation relating P and Q.

aev [14]

P= k/Q
16.4= k/123
k= 16.4 * 123
k= 2017.2
Therefore, P= 2017.2/Q

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4 years ago

Derive the equation of the parabola with a focus at (3,1) and a directrix of y = 5

serg [7]

So hmmm check the picture below

so... the vertex is "p" distance from the focus and the directrix, thus, the vertex is really half-way between both

in this case, 2 units up from the focus or 2 units down from the directrix, and thus it lands at 3,3

now, the "p" distance is 2, however, the directrix is up, the focus point is below it, the parabola opens towards the focus point, thus, the parabola is opening downwards, and the squared variable is the "x"

because the parabola opens downwards, "p" is negative, and thus, -2

now, let's plug all those fellows in then

$\bf \begin{array}{llll}
(x-{{ h}})^2=4{{ p}}(y-{{ k}})\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-----------------------------\\\\

\begin{cases}
h=3\\
k=3\\
p=-2
\end{cases}\implies (x-3)^2=4(-2)(y-3)\implies (x-3)^2=-8(y-3)
\\\\\\
-\cfrac{(x-3)^2}{8}=y-3\implies \boxed{-\cfrac{1}{8}(x-3)^2+3=y}$