What’s the product of 1,243x8

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:

Yes

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:

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:

the line perpendicular: x=-3

Step-by-step explanation:

Answer:

$90

Step-by-step explanation:

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:

4/10, 5/10, 6/10

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

Answer:

(5)

for eg;

3/5 is proportional to 15/25

Answer:

I would say 1 or 4

Step-by-step explanation:

Proportional relationships are relationships between two variables where their ratios are equivalent. Another way to think about them is that, in a proportional relationship, one variable is always a constant value times the other. That constant is know as the "constant of proportionality".

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:

-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 answer is 1/2 second

Step-by-step explanation:

Hope this helps

Answer:

y=5x-3

Step-by-step explanation:

it just is

Answer:

y=5x-3

Step-by-step explanation:

Step-by-step explanation:

i am 90% sure this is correct

Answer:

Step-by-step explanation:

Do multiplication first since it comes before the division in this equation

1/5 · 1/2 = 1/10

Plug this value back into the equation

1/2 - 1/10 + 2/5 ÷ 7/10

Divide 2/5 by 7/10

4/7

Plug this value back into the equation

1/2 - 1/10 + 4/7

Subtract 1/10 from 1/2

2/5

Plug this value back into the equation

2/5 + 4/7

Add.

34/35

Hope this helps :)

Please consider Brainliest :)

Answer:

whith whut

Step-by-step explanation:

Answer:

6 miles

Step-by-step explanation:

3 minutes = 1/20 of an hour

120 miles x 1/20 of an hour = 6 minutes

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:

35

Step-by-step explanation:

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 recent role of the greenhouse effect

Step-by-step explanation:

I believe you answer would be D) The distance traveled by a truck driving at a constant speed compared with time.

Step-by-step explanation:

well that seems like a you problem lol

Answer:

x= 1/5 or -2

Step-by-step explanation:

5x² +11x +2= 0

(Divide through by the coefficient of x²)

5x²/5 + 11x/5 + 2/5 = 0/5

x² + 11x/5 + 2/5 = 0

(take the constant to the other side)

x² + 11x/5 = -2/5

(Add 1/2 of the coefficisint of x all square to both sides)

x² + x + (11/5 × 1/2)² = -2/5 + (11/5 × 1/2)²

x² + x + (11/10)² = -2/5 + (11/10)²

x² + x + (11/10)² = -2/5 + 121/100

(eliminate x )

x² + (11/10)² = (20(-2) + 121 ) ÷ 100

(x + 11/10 )² = (-40 + 121) ÷ 100

(x + 11/10 )² = 81/100

(square both sides)

x + 11/ 10 =√(81/100)

x + 11/10 = 9/10

x = -11/10 +or- 9/10

x = -11/10 + 9/10 or -11/10 - 9/10

x = (-11 + 9) ÷ 10 or (-11-9) ÷ 10

x = -2/ 10 or -20/10

x = -1/5 or -2

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]