Anwser to all the questions please

The equation for the line of best fit is y=2x+2.5. Soon, Edmond is going to a new fair. Use the equation for the line of best fi

andriy [413]

Answer:

x = 5

Step-by-step explanation:

The equation of a line is given by :

y=2x+2.5

Where

y is cost and x is number of tickets

Put y = $12.5 in the above equation

So,

12.5=2x+2.5

10 = 2x

x = 5

So, he can purchase 5 tickets.

4 0

3 years ago

the equation is -2x² = 4-3 (x + 1) and the question is justify that it is a 2nd degree equation with the unknown x complete.

Serggg [28]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the equation

-2x² = 4-3 (x + 1)

-2x² = 4-3x-3

-2x² = -3x -7

0 = 2x² -3x -7

We know that the degree of the equation is the highest power of x variable in the given equation.

In the equation 0 = 2x² -3x -7 the highest power of x variable in the given equation is 2.

Thus, the degree of the equation is 2.

Also in the equation 0 = 2x² -3x -7, the unknown variable is 'x'.

Let us determine the value 'x'

2x² -3x -7 = 0

Add 7 to both sides

$2x^2-3x-7+7=0+7$

$2x^2-3x=7$

Divide both sides by 2

$\frac{2x^2-3x}{2}=\frac{7}{2}$

$x^2-\frac{3x}{2}=\frac{7}{2}$

Add (-3/4)² to both sides

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{7}{2}+\left(-\frac{3}{4}\right)^2$

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{65}{16}$

$\left(x-\frac{3}{4}\right)^2=\frac{65}{16}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{3}{4}=\sqrt{\frac{65}{16}}$

$x-\frac{3}{4}=\frac{\sqrt{65}}{\sqrt{16}}$

$x-\frac{3}{4}=\frac{\sqrt{65}}{4}$

Add 3/4 to both sides

$x-\frac{3}{4}+\frac{3}{4}=\frac{\sqrt{65}}{4}+\frac{3}{4}$

$x=\frac{\sqrt{65}+3}{4}$

similarly solving

$x-\frac{3}{4}=-\sqrt{\frac{65}{16}}$

$x=\frac{-\sqrt{65}+3}{4}$

So the solution of the equation will have the values of x such as:

$x=\frac{\sqrt{65}+3}{4},\:x=\frac{-\sqrt{65}+3}{4}$

6 0

3 years ago

Amanda had $20 when she went into the shopping center. She purchased a snack to eat for $y and then bought two items at a clothi

Volgvan

Answer:

x+y=20

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

(30 points) Enter numbers to write 0.000328 in scientific notation.

Sloan [31]

$0.000328= \cfrac{3.28}{10,000}=3.28*10^{-4}$

4 0

3 years ago

Consider all 5 letter "words" made from the full English alphabet. (a) How many of these words are there total? (b) How many of

VARVARA [1.3K]

Answer:

a) There are 11,881,336 of these words in total.

b) There are 7,893,600 of these words with no repeated letters.

c) 896,376 of these words start with an a or end with a z or both

Step-by-step explanation:

Our words have the following format:

L1 - L2 - L3 - L4 - L5

In which L1 is the first letter, L2 the second letter, etc...

There are 26 letters in the English alphabet.

(a) How many of these words are there total?

Each of L1, L2, L3, L4 and L5 have 26 possible options.

So there are $26^{5} = 11,881,336$ of these words total

(b) How many of these words contain no repeated letters?

The first letter can be any of them, so L1 = 26.

At the second letter, the first one cannot be repeated, so L2 = L1 - 1 = 25.

At the third letter, nor the first nor the second one can be repeated, so L3 = L1 - 2 = 24

This logic applies until L5

So we have

26-25-24-23-22

In total there are

$26*25*24*23*22 = 7,893,600$

of these words with no repeated letters.

(c) How many of these words start with an a or end with a z or both (repeated letters are allowed)?

$T = T_{1} + T_{2} + T_{3}$

$T_{1}$ is the number of words that start with an a and do not end with z. So we have

1 - 26 - 26 - 26 - 25

The first letter can only be a, and the last one cannot be z. So:

$T_{1} = 26^{3}*25 = 439,400$

$T_{2}$ is the number of words that start with any letter other than a and end with z. So we have

25 - 26 - 26 - 25 - 1

The first letter can be any of them, other than a, and the last can only be z. So:

$T_{2} = 26^{3}*25 = 439,400$

$T_{3}$ is the number of words that both start with a and end with z. So:

1 - 26 - 26 - 26 - 1

The first letter can only be a, and the last can only be z. The other three letters could be anything. So:

$T_{3} = 26^{3} = 17,576$

$T = T_{1} + T_{2} + T_{3} = 2*439,400 + 17,576 = 896,376$

896,376 of these words start with an a or end with a z or both

4 0

3 years ago