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3 years ago

Graph the equation y=x^2-1

Mathematics

1 answer:

TiliK225 [7]3 years ago

7 0

Answer below:

Hope this helps :-)

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algol13

2 - 2 x 3 + 3

= 2 - 6 + 3

= -4 + 3

= - 1

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4 years ago

How many Solutions does 5p + 3 = 5p - 1 have?

Tresset [83]

No solution because 3=-1 isn’t true.

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3 years ago

Please help me with 19 and 20 please

lesya [120]

9514 1404 393

Answer:

19. B -- continued, but modest ...

Step-by-step explanation:

19. There is no decline or decrease indicated on this graph. If growth were exponential, the graph would be concave upward, which it is not. There is continued growth indicated.

20. The percentage change from 2005 to 2010 is ...

(60 -20)/20 × 100% = 2 × 100% = 200%

One might compute an average rate of change per year of ...

200%/(5 yr) = 40%/yr

_____

Additional comment

As with any statement of percentage, you need to be very clear about what the base is.

Here, 100% is the number of farms in 2005, so an increase of 40% per year is an increase by 40% of the number in 2005. That is very different from 40% of the number in the previous year, which is how an annual percentage increase is usually interpreted. (The average annual rate of change is closer to 24% with respect to the previous year's number.)

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3 years ago

(1) The Yearling is one of my favorite novels.(2) It is by Marjorie Kinnan Rawlings. (3) It is about a yearling, which is a youn

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7 0

3 years ago

Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).

andre [41]

Check the picture below. So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

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

$\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}$

7 0

3 years ago