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10 months ago

use the graphing calculator to graph the quadratic function y = x2 − 5x − 36. Which value is a solution of 0 = x2 − 5x − 36?

Mathematics

1 answer:

daser333 [38]10 months ago

3 0

The solutions of the quadratic equation x² - 5x - 36 = 0 are -4 and 9.

The graph is attached below.

We are given a quadratic function.
A polynomial equation of degree two in one variable is a quadratic equation.
The function given to us is :
y = x² - 5x - 36
We need to find the solution of the quadratic function.
To find the roots, let y = 0.
x² - 5x - 36 = 0
Use the quadratic formula.
In elementary algebra, the quadratic formula is a formula that gives the solution(s) to a quadratic equation.
x = [-b±√b²-4ac]/2a
x = [-(-5) ± √25 - 4(1)(-36)]/2(1)
x = (5 ± √25 + 144)/2
x = (5 ± √169)/2
x = (5 ± 13)/2
x = 9 or x = -4

To learn more about functions, visit :

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2 years ago

Two numbers have a product of 20. The second number is three less than twice the first number. What are the two numbers?

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2 years ago

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Maksim231197 [3]

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3 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

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2 years ago

At the park, there is a triangular sandbox with base = 12ft and height = 4ft. A drawing of the sandbox is shown above. Brie want

sammy [17]

Your question is not clear, but it looks as though you want to know how Brie can make a similar sandbox with base = 8ft.

Answer:

For Brie to make a similar sandbox, he must use a base = 8ft, and height = (8/3)ft

Step-by-step explanation:

It is possible for Brie to make a similar triangular sandbox with base = 12ft and height = 4ft.

All he must ensure is that the ratio of base to height of the original sandbox is the same ratio of base to height of the one he is trying to make.

The original sandbox is 12:4

Because he wants to use a base = 8ft, the sandbox he is trying to make is 8:x

Where x is the height of the sandbox he is trying to make.

Then for these sandboxes to be similar, the ratio 12:4 = 8:x

=> 12/4 = 8/x

12x = 8 × 4

12x = 32

x = 8/3

The height must be (8/3)ft

3 0

2 years ago