In the case of a quadratic equation ax2 + bx + c = 0, the discriminant is b2 − 4ac; for a cubic equation x3 + ax2 + bx + c = 0, the discriminant is a2b2 + 18abc − 4b3 − 4a3c − 27c2.

Step-by-step explanation:

4 0

3 years ago

Write a equation of a hyperbola given the foci and the asymptotes

professor190 [17]

Solution:

The standard equation of a hyperbola is expressed as

$\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\text{ \lparen parallel to the x-axis\rparen} \\ \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\text{ \lparen parallel to the y-axis\rparen} \end{gathered}$

Given that the hyperbola has its foci at (0,-15) and (0, 15), this implies that the hyperbola is parallel to the y-axis.

Thus, the equation will be expressed in the form:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\text{ ----equation 1}$

The asymptote of n hyperbola is expressed as

$y=\pm\frac{a}{b}(x-h)+k$

Given that the asymptotes are

$y=\frac{3}{4}x\text{ and y=-}\frac{3}{4}x$

This implies that

$a=3,\text{ and b=4}$

To evaluate the value of h and k,

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3 0

1 year ago

What is the range of the function y=2x+1 for the domain 2<=x<=5?

egoroff_w [7]

So domain is the number you can use
range is the output your get from inputting the domain given
so from 2≤x≤5
since it is linear, we can be sure that we only need to test the endpoints of the domain to find the endpoints of the range

sub 2 for x
y=2(2)+1
y=4+1
y=5

sub 5 for x
y=2(5)+1
y=10+1
y=11

so range is from 5 to 11
in interval notation: [5,11]
in other notation 5≤y≤11
or
R={y|5≤y≤11}

5 0

3 years ago

PLZ QUICK!!

iragen [17]

Answer:

See attached graph

Step-by-step explanation:

The equation is y-3=(x + 6)

Write the equation in a slope intercept form, then graph the equation on a graph tool to see the points that line on the line.

Alternatively , using the coordinates in the answer choices given, input them in the equation of the graph and select the answer choice that has all its coordinates true to the equation.

y-3 = x+6 -----can be written as : y= x+9

Graph y= x+9 to see the graph and the points that are on the line as attached.

3 0

3 years ago

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