The sum of three consecutive integers is 9.

Find the x-intercepts of the parabola with vertex (1,1) and y-intercept (0,-3). Write your answer in this form (x of 1 , y of 1)

Vikki [24]

Remember that the vertex form of a parabola or quadratic equation is:

y=a(x-h)^2+k, where (h,k) is the "vertex" which is the maximum or minimum point of the parabola (and a is half the acceleration of the of the function, but that is maybe too much :P)

In this case we are given that the vertex is (1,1) so we have:

y=a(x-1)^2+1, and then we are told that there is a point (0,-3) so we can say:

-3=a(0-1)^2+1

-3=a+1

-4=a so our complete equation in vertex form is:

y=-4(x-1)^2+1

Now you wish to know where the x-intercepts are. x-intercepts are when the graph touches the x-axis, ie, when y=0 so

0=-4(x-1)^2+1 add 4(x-1)^2 to both sides

4(x-1)^2=1 divide both sides by 4

(x-1)^2=1/4 take the square root of both sides

x-1=±√(1/4) which is equal to

x-1=±1/2 add 1 to both sides

x=1±1/2

So x=0.5 and 1.5, thus the x-intercept points are:

(0.5, 0) and (1.5, 0) or if you like fractions:

(1/2, 0) and (3/2, 0) :P

4 0

3 years ago

Please help, seriously confused

Rufina [12.5K]

Original height = original width = x mm

1. She made an enlarged copy...
height = 2x
width = 2x

2. She cut off a rectangle...
height = 2x
width = $\frac{2}{3} *2x= \frac{4}{3}x$

3.She doubled the width...
height = 2x
width = $2*\frac{4}{3}x = \frac{8}{3}x$

height * width = area

$2x * \frac{8}{3}x=139 \ 968 \\ \\ \frac{16}{3}x^2= 139 \ 968 \\\\ x^2=139 \ 968: \frac{16}{3}=139 \ 968 * \frac{3}{16}= 8 \ 748*3 = 26 \ 244 \\ \\ x= \sqrt{26 \ 244} = 162 \ mm$

Original height was 162 mm

6 0

3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

2 years ago

What is the quotation of 7^-1/7^2

Temka [501]

Answer:

7^-3

Step-by-step explanation:

<u>Solving in steps</u>

7^-1/7^2 =
7^-1 × 7^-2 =
7^(-1 - 2) =
7^-3

3 0

3 years ago

4 green marbles, 5 red marbles, and 2 yellow marbles are placed into a bag. One marble is chosen and replaced into the bag. A se

Ipatiy [6.2K]

Answer:

The probability is 10/121

Step-by-step explanation:

In this question, we are tasked with calculating probability.

The probability to be calculated is that we have an ordered selection of picking a red marble after which we pick a yellow marble.

Firstly let us know the total number of marbles we have. That will be 4 + 5 + 2 = 11 marbles

Probability of picking a red marble will be 5/11 while the probability of picking a yellow marble will be 2/11

Now, the probability of picking a red marble before a yellow marble will be mathematically equal to; P(r) * P(y) = 5/11 * 2/11 = 10/121

7 0

2 years ago