All categories

3 years ago

30th term: a = 4, d = 12

Mathematics

1 answer:

mezya [45]3 years ago

8 0

Answer:

120

Step-by-step explanation:

4*30=120

You might be interested in

Which statement is Correct

mars1129 [50]

Answer:

answer : C

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

1. Expresa cada función cuadrática f en la forma f(x)-yo = p(x – xo) x ϵ R, donde xo , yo , p ϵ R son elegidos apropiadamente. I

crimeas [40]

Answer:

Step-by-step explanation:

1. Express each quadratic function f in the form f (x) -yo = p (x - xo) x ϵ R, where xo, i, p ϵ R are appropriately chosen. Indicate the coordinates

of the vertex of the parabola, the minimum or maximum of the function f, especially R. Draw the graph of f.

a) f (x) = 2x2 + 3x -5, ϵ R,

b) f (x) = -3x2 + 7x +9, ϵ R,

c) f (x) = - x +1, ϵ R,

HELP YOU GIVE YOU CORONA PLIS

7 0

3 years ago

Midpoint of (-2,-1) (-6,12)

Leni [432]

Answer:

(-4, 5.5)

Step-by-step explanation:

1. put your Xs and Ys over 2

2. add them together

3. Divide

4. Simplify

5. Turn fraction into decimal

$m(x.y) \\ m = \frac{x1 + x2}{2} . \: \frac{y1 + y2}{2} \\ m = \: \frac{ - 2 + - 6}{2} . \frac{ - 1 + 12}{2} \\ m = ( - \frac{8}{2} . \frac{11}{2} ) \\ m = ( - 4 \: . \: 5 \frac{1}{2} ) \\ \\ \\ mid = ( - 4 \: \: . \: 5.5)$

6 0

2 years ago

A triangle has one side length of 2x

Orlov [11]

The simplified expression that represents the total perimeter of the triangle is 2x^2 + 10x + 3

<h3>How to determine the perimeter?</h3>

The side lengths are given as:

2x^2 + 5x + 3, 2 + 2x and 3x - 2

Add these lengths to determine the perimeter (P)

P = 2x^2 + 5x + 3 + 2 + 2x + 3x - 2

Collect like terms

P = 2x^2 + 5x + 2x + 3x + 3 + 2 - 2

Evaluate the like terms

P = 2x^2 + 10x + 3

Hence, the simplified expression that represents the total perimeter of the triangle is 2x^2 + 10x + 3

Read more about perimeter at:

brainly.com/question/6465134

#SPJ1

8 0

2 years ago

How can i differentiate this equation?

Dmitry_Shevchenko [17]

$\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}$

$\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}$

$\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x\sqrt{x}} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{\sqrt{x}}$

8 0

3 years ago