a bus is full of passengers. if you count them by either twos, threes, or fives, there is one left. if you count them by seven there will be none left. find the least number of passengers in the bus?​

Answer:

-33

Step-by-step explanation:

I hope this is correct, and I apologize in advance if it isn't :(

Answer:

g(-3) = -33

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Step-by-step explanation:

Step 1: Define

g(x) = 2x³ + 2x² - 4x - 9

g(-3) is x = -3

Step 2: Evaluate

Answer:

The recent role of the greenhouse effect

Step-by-step explanation:

Answer:

1.222

Step-by-step explanation:

Answer:

305m+152

Step-by-step explanation:

(8m+4)38

Multiply each term in the parentheses

8m times 38 plus 4 times 38

Multiply 8m times 38

304m plus 4 times 38

Then Multiply 4 times 38

304m+152

That should be your answer, but I could be wrong, therefore I am so sorry if I am. I hope this helps you.

Answer:

21

Step-by-step explanation:

I believe this because when you add it all together you get 20.77

Answer:

x=4

Step-by-step explanation:

Answer:

(n+10)+9

{(n+10)+9}/2

[{(n+10)+9}/2]-5

All you have to do is follow the steps!

Think of a number - n

Add 10 - n + 10

Add the number you were thinking of - n + 10 + n

Divide by 2 - (n + 10 + n) / 2

Subtract 5 - ((n + 10 + n) / 2) -5

It may look hard, but its not as hard as you think. All you need to do is follow the steps!

Answer:

Age of Anna = x years

Age of Jane i.e. Anna's sister = (x-y) years

a, What was the sum of their ages two years ago?

Solution,

Age of Anna two years ago = (x-2) years

Age of Jane two years ago = (x-y-2) years

Sum of their ages = x - 2 + x-y-2 = 2x - y -4

b. What will be the sum of their ages in 3 years?

Solution,

We have,

Present age of Anna = x years

Present age of Jane = (x-y) years

So,

Age of Anna after 3 years = (x+3) years

Age of Jane after 3 years = (x-y+3) years

Therefore,

Sum of their ages in 3 years = (x+3) +(x-y+3) = 2x - y+6

c. How old was Anna when Jane was born?

Solution,

Here,

According to question,

Jane is y years younger than Anna

So, Anna was y years when Jane was born.

[Suppose your age is 10 years and your brothers age is 14 years. You are 4 years younger than your brother. Also, his age was 4 years when you were born.]

Answer:

Yes

Step-by-step explanation:

Answer:

whith whut

Step-by-step explanation:

Answer:

24x+18y+10

Step-by-step explanation:

start by distributing so 6*4x and 6*3y

that give you 24x+18y+10

Answer:

35

Step-by-step explanation:

Answer:

Step-by-step explanation:

Use the distributive property to get rid of the brackets.

9i + 2*7 + 2*3i        Use the multiplication property to simplify the imaginary terms

9i + 2*3i + 2*7        Do the same to the real term

9i + 6i + 14              Use the addition of like terms property

15i + 14                    And that's your answer.

You may not like my terminology, but you have to remember that every text teaches this differently.

Answer:

[tex]y''=\frac{-250x}{y^5}[/tex]

Choice A.

Step-by-step explanation:

We are given:

[tex]5x^3-y^3=25[/tex]

We are asked to find:

[tex]\frac{d^2y}{dx^2}[/tex]

This requires we first find:

[tex]\frac{dy}{dx}[/tex]

Let's begin.

We will differentiate the right hand side using the constant rule since 25 is just a constant. We will differentiate the left hand side using difference rule  and then constant multiple rule and then power rule and chain rule.

[tex](5x^3-y^3=25)'[/tex]

[tex](5x^3-y^3)'=(25)'[/tex]

[tex](5x^3)'-(y^3)'=0[/tex]

[tex]5(x^3)'-(y^3)'=0[/tex]

[tex]15x^2-3y^2y'=0[/tex]

Now let's solve for [tex]y'[/tex].

Add [tex]3y^2y'[/tex] on both sides:

[tex]15x^2=3y^2y'[/tex]

Divide both sides by [tex]3y^2[/tex]:

[tex]\frac{15x^2}{3y^2}=y'[/tex]

Symmetric property of equality:

[tex]y'=\frac{15x^2}{3y^2}[/tex]

Simplify right hand side:

[tex]y'=\frac{5x^2}{y^2}[/tex]

Now let's find the second derivative by differentiating both sides of this equation. We will use the quotient rule on the right hand side:

[tex]y''=\frac{(5x^2)'y^2-5x^2(y^2)'}{(y^2)^2}[/tex]

[tex]y''=\frac{10xy^2-5x^2(2yy')}{y^4}[/tex]

We can replace [tex]y'[/tex] with [tex]\frac{5x^2}{y^2}[/tex]... since we just got this result before differentiating both sides of an equality again.

So we have:

[tex]y''=\frac{10xy^2-5x^2(2y\frac{5x^2}{y^2})}{y^4}[/tex]

Let's simplify the 2nd term on top by cancelling a common factor of y in that term from top and bottom:

[tex]y''=\frac{10xy^2-5x^2(2\frac{5x^2}{y})}{y^4}[/tex]

We still don't like having this compound fraction (a mini fraction inside a bigger fraction). Let's get rid of that...Multiply numerator and denominator by [tex]\frac{y}{y}[/tex] for the big fraction to clear the mini fraction out.

[tex]y''=\frac{y}{y}\frac{10xy^2-5x^2(2\frac{5x^2}{y})}{y^4}[/tex]

Distribute:

[tex]y''=\frac{10xy^3-5x^2(2)5x^2}{y^5}[/tex]

Simplifying a bit more:

[tex]y''=\frac{10xy^3-50x^4}{y^5}[/tex]

We could factor the top a bit... both terms on top of a common factor of [tex]10x[/tex]:

[tex]y''=\frac{10x(y^3-5x^3)}{y^5}[/tex]

Notice the thing in ( ) on top is the exact opposite of the left hand side of our original equation: [tex]5x^3-y^3=25[/tex].

This implies we can replace it with -25:

[tex]y''=\frac{10x(-25)}{y^5}[/tex]

[tex]y''=\frac{-250x}{y^5}[/tex]

Answer:

The answer is 1/2 second

Step-by-step explanation:

Hope this helps

Step-by-step explanation:

well that seems like a you problem lol

Answer:

4/10, 5/10, 6/10

I'm guessing 6/10 simplified into 3/5

Answer:

Step-by-step explanation:

x = 0

x+3 = 0

x = -3

The answer is x could be 0 or -3

Answer:

c on edge

Step-by-step explanation:

Answer:

Step-by-step explanation:

Step one:

given data

First tree

the tree is 8cm tall

and 30cm from the vanishing point

Third  tree

let the tree be x cm tall

and 20cm from the vanishing point

Step two:

if 8cm tree will be 30cm at the vanishing point

then x cm tree will be 20cm at the vanishing point

cross multiply we have

x= (20*8)/30

x=160/30

x=5.3cm

The third tree is 5.3cm tall

Answer:

y=5x-3

Step-by-step explanation:

it just is

Answer:

y=5x-3

Step-by-step explanation:

Answer:

The answer is "np(1+p)".

Step-by-step explanation:

[tex]N_1[/tex] = Number of n tossed heads  

According to [tex]N_1[/tex], Y = amount of [tex]N_1[/tex] tossing heads

So, [tex]N_1[/tex]~ Bin(n, p) and [tex]N_2[/tex] | [tex]N_1[/tex] ~ Bin([tex]N_1[/tex], p)

[tex]E(N_1+N_2) & = E\Big[N_1+N_2\vert N_1\Big]\\ &[/tex]

                   [tex]= E(N_1)+E\Big[N_2\vert N_1\Big]\\\\ & = E(N_1) + E(N_1\,p)\\\\ & = np + np.p \\\\ =np+np^2\\\\ =np(1+p)[/tex]

Explanation:

Remember, the term analytics (or data analytics) refers to the act of carefully examining raw data in other to observe if there are any trends.

a) The following are three cases in our daily life/work where analytics could be applied:

b) For example, to analyze changes in my sleeping patterns based on hours spent on electronic gadgets, I need to first know

Answer:

complement of what ?

Step-by-step explanation:

Answer:

[tex](7,-11)[/tex]

Step-by-step explanation:

Given

[tex]S = (1,5)[/tex]

[tex]M = (4,-3)[/tex]

Required

Determine the coordinates of T

Midpoint is calculated as:

[tex]M(x,y) = \frac{1}{2}(x_1 +x_2,y_1+y_2)[/tex]

Where:

[tex](x,y) = (4,-3)[/tex]

[tex](x_1,y_1) = (1,5)[/tex]

Substitute these values in the above formula:

[tex](4,-3) = \frac{1}{2}(1 + x_2,5+y_2)[/tex]

Multiply through by 2

[tex]2 * (4,-3) = 2 * \frac{1}{2}(1 + x_2,5+y_2)[/tex]

[tex](8,-6) = (1 + x_2,5+y_2)[/tex]

By direct comparison:

[tex]1 + x_2 = 8[/tex] and [tex]5 + y_2 = -6[/tex]

Solving for x2

[tex]1 + x_2 = 8[/tex]

[tex]x_2 = 8 - 1[/tex]

[tex]x_2 = 7[/tex]

Solving for y2

[tex]5 + y_2 = -6[/tex]

[tex]y_2 = -6 - 5[/tex]

[tex]y_2 = -11[/tex]

Hence, the coordinates of T is: [tex](7,-11)[/tex]

Answer:

The gradient of the graph below is [tex]\mathbf{\frac{3}{2} }[/tex]

Step-by-step explanation:

We need to calculate the gradient of the graph

The gradient of graph is actually slope of the graph.

The slope of graph can be calculated using formula: [tex]Slope=\frac{y_2-y_1}{x_2-x_1}[/tex]

Taking any 2 points on graph i.e

(1,0) and (-1,-3)

We have [tex]x_1=1, y_1=0, x_2=-1,y_2=-3[/tex]

Putting values and finding gradient:

[tex]Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{-3-0}{-1-1}\\Slope=\frac{-3}{-2}\\Slope=\frac{3}{2}[/tex]

So, The gradient of the graph below is [tex]\mathbf{\frac{3}{2} }[/tex]

Answer:

A

Step-by-step explanation:

Unit Test on a Edge today. ; )

Answer:

its A

Step-by-step explanation:

i got a 100  on my quiz