Did I get this right if not can you help me

What is true about the dilation?

OlgaM077 [116]

Answer:

<h2>pretty sure its A sorry if its wrong </h2>

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A small box in the shape of a cube for packaging has a volume of 64 cubic inches.

Oduvanchick [21]

The volume of a medium box is 512 cubic inches.
The ratio of the sides of the small box to the medium box is 1:2.
The ratio of the area of the small box to the medium box is 1:4.
The ratio of the volume of the small box to the medium box is 1:8.

<h3>What are the ratio of the small box to the medium box?</h3>

The first step is to determine the side lengths of the small box.

Side length = ∛64 = 4 in

Side lengths of the medium boxes = 4 x 2 = 8 inches

Volume of the medium box = 8³ = 256 cubic inches

The ratio of the sides of the small box to the medium box = 4 : 8 = 1:2.

The ratio of the area of the small box to the medium box= (4 x 4) : (8 x 8) = 1:4.

The ratio of the volume of the small box to the medium box = 4³ : 8³ = 1 : 8.

To learn more about the volume of a cuboid, please check: brainly.com/question/26406747

5 0

2 years ago

The x-coordinate of the vertex of the equation y = 2x2 − 4x + 12 The x-coordinate of the vertex of the equation y = 4x2 + 8x + 3

dolphi86 [110]

So since the vertex falls onto the axis of symmetry, we can just solve for that to get the x-coordinate of both equations. The equation for the axis of symmetry is $x=\frac{-b}{2a}$ , with b = x coefficient and a = x^2 coefficient. Our equations can be solved as such:

y = 2x^2 − 4x + 12: $x=\frac{4}{2*2}=\frac{4}{4}=1$

y = 4x^2 + 8x + 3: $x=\frac{-8}{2*4}=\frac{-8}{8}=-1$

In short, the vertex x-coordinate's of y = 2x^2 − 4x + 12 is 1 while the vertex's x-coordinate of y = 4x^2 + 8x + 3 is -1.

6 0

3 years ago

A 95% confidence interval for an unknown population mean is given by (9, 12). How should this confidence interval be interpreted

Zanzabum

C is my answer of ask other sources tho. Ok

8 0

2 years ago

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

Set <em>u</em>: $\displaystyle u = 4x^{-2}$
[<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

<u>Step 3: Integrate Pt. 2</u>

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

4 0

2 years ago