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3 years ago

What are the zeroes of f(x)=(x-1)(x+1)?

Mathematics

1 answer:

Leni [432]3 years ago

6 0

Answer:

-1 and 1

Step-by-step explanation:

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A candy company claims that its jelly bean mix contains 15% blue jelly beans. Suppose that the candies are packaged at random in

I am Lyosha [343]

Answer:

Doubtful.

Step-by-step explanation:

Our values are,

$P=0.15 \rightarrow 15\%$ Jelly Bean Mix.

The sample for us is n=200

And our proportion (x) is 40 .

We can calculate with this dates the sample proportion and make he test statistic.

$p=\frac{x}{n} = \frac{40}{200} = 0.20$

$H_0:P=0.15$

$H_1:P \neq 0.15$

We use the formula for Test statistic, that is given by,

$z= \frac{p-P}{\sqrt{\frac{P-(1-P)}{n}}}$

$z= \frac{0.20-0.15}{\sqrt{\frac{0.15(1-0.15)}{200}}}$

$z= 1.98$

We see that our P-value: 0.048 (From the Table)

We note that p-value is less than 0.05, so we should to rejecyt the null hypothesis and conclude that the calim is doubtful

4 0

3 years ago

The volume of a cube of ice for an ice sculpture is 64000

-BARSIC- [3]

no question to answer

Step-by-step explanation:

8 0

3 years ago

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago

What does Inductory mean

timofeeve [1]

Serving to induct or bring in; introductory

8 0

3 years ago

Write an equation for the line perpendicular to the line y= -2/3+7 which passes through the point (-5,6)

zysi [14]

Y = -2/3x + 7.....the slope here is -2/3. A perpendicular line will have a negative reciprocal slope. All tht means is take the original slope, flip it, and change the sign. So we take -2/3....flip it making it -3/2.....change the sign making it 3/2. So ur perpendicular line will have a slope of 3/2.

y = mx + b
slope(m) = 3/2
(-5,6)....x = -5 and y = 6
now sub and find b, the y int
6 = 3/2(-5) + b
6 = -15/2 + b
6 + 15/2 = b
12/2 + 15/2 = b
27/2 = b
so ur perpendicular equation is : y = 3/2x + 27/2 <==

8 0

3 years ago