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vodka [1.7K]
3 years ago
9

How to determine the function with the graph?

Mathematics
1 answer:
Roman55 [17]3 years ago
3 0
I think it’s f (x)= 1
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Help please! Could you explain it too?
Anettt [7]

Answer:

a) not proportional

b) proportional; k = -\frac{3}{5}

Step-by-step explanation:

a) for any proportional equation, the line must pass through the origin. The equation in a) is y = 4x + 1, and the '+1' is the y-intercept. This means that the line does not pass through the origin, so x and y cannot increase by the same amount (i.e. they are not proportional).

Another way to determine this is is to use the y = kx base. If you have an equation that fits that it's likely proportional.

Here, if the equation was only y = 4x then it'd be proportional because u can see that k = 4. This is not the equation though, and the 4x + 1 doesn't fit to the y = kx formula so it can't be proportional.

b) straight away you can see that there's no 'c' term (y = mx + c) which means the y-intercept is 0, so the line passes through the origin. While this does not immediately mean the line is proportional, you can make sure that it is by checking it fits with the y = kx equation.

y = -(3/5)x fits with y = kx, with k being -3/5

3 0
2 years ago
Write an equation of the hyperbola given that the center is at (2, -3), the vertices are at (2, 3) and (2, - 9), and the foci ar
zavuch27 [327]
Check the picture below.

so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).

because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,

\bf \textit{hyperbolas, vertical traverse axis }
\\\\
\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1
\qquad 
\begin{cases}
center\ ( h, k)\\
vertices\ ( h,  k\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2 + b ^2}\\
asymptotes\quad  y= k\pm \cfrac{a}{b}(x- h)
\end{cases}\\\\
-------------------------------

\bf \begin{cases}
h=2\\
k=-3\\
a=6\\
c=2\sqrt{10}
\end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1
\\\\\\
\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1
\\\\\\
c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2
\\\\\\
4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2
\\\\\\
\sqrt{4}=b\implies 2=b\\\\
-------------------------------\\\\
\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1

3 0
2 years ago
a sequence starts a 200 and 30 is subtracted each time 200,170,140 what are the first two numbers in the sequence that are kess
Snowcat [4.5K]
<h3>Answer:  -10 and -40</h3>

===============================================================

Explanation:

a = 200 = first term

d = -30 = common difference

Tn = nth term

Tn = a + d(n-1)

Tn = 200 + (-30)(n-1)

Tn = 200 - 30n + 30

Tn = -30n + 230

Set Tn less than 0 and isolate n

Tn < 0

-30n + 230 < 0

230 < 30n

30n > 230

n > 230/30

n > 7.667 approximately

Rounding up to the nearest whole number gets us n \ge 8

So Tn starts to turn negative when n = 8

We can see that,

Tn = -30n + 230

T7 = -30*7 + 230

T7 = 20

and

Tn = -30n + 230

T8 = -30*8 + 230

T8 = -10 is the 8th term

and lastly

Tn = -30n + 230

T9 = -30*9 + 230

T9 = -40 is the ninth term

Or once you determine that T7 = 20, you subtract 30 from it to get 20-30 = -10 which is the value of T8. Then T9 = -40 because -10-30 = -40.

5 0
3 years ago
Read 2 more answers
What is 1cm of 1 metre as a fraction?
777dan777 [17]
1/100 of a metre, as 100cm make 1m
6 0
3 years ago
Read 2 more answers
The density of a substance is the mass of the substance per unit of volume. The density of the element Americium is inversely pr
Viktor [21]

Answer:

(3,4)

Step-by-step explanation:

7 0
2 years ago
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