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2 years ago

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A photographer made a 5-inch by 8-inch print of a photo. He then made a 10-inch by 16-inch print of the same photo. Which scale

factor was use to create the larger photo from the smaller photo?
A. 2

B. 4

C. 5

D. 6

E. 8

1 answer:

torisob [31]2 years ago

7 0

A.2 because 8*2 =16 and 5*2=10

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Amanda [17]

Answer:

1. $1200 \ kg/m^3$

2.

a)

35 cm = 0.35 meters

11 dm = 1.1 meters

15 mm = 0.015 meters

b)

Volume = 0.005775 cubic meters

Mass = 15.5925 kilograms

Step-by-step explanation:

1.

We know

Density = Mass/Volume

Given

Mass = 90 kg

Volume = 0.075 cubic meters

The SI unit for density is kilograms per cubic meters. Our dimensions are just that, so we just need to plug in the numbers into the formula and get our answer. Shown below:

Density = $\frac{90}{0.075}=1200 \ kg/m^3$

The Density of asphalt block = $1200 \ kg/m^3$

2.

a)

We need to express 35cm, 11dm, and 15mm in meters

We know

100 cm = 1 meters, so

35 cm = 35/100 = 0.35 meters

We know 1 dm = 0.1 meters, so

11 dm * 0.1 = 1.1 meters

We know, 1mm = 0.001 meters, so

15 mm * 0.001 = 0.015 meters

b)

Volume of the Slab is length * width * height, in meters, that would be:

Volume = 0.35 * 1.1 * 0.015 = 0.005775 cubic meters

THe mass would be found by the formula:

Density = Mass/Volume

2700 = Mass/0.005775

Mass = 0.005775 * 2700 = 15.5925 kilograms

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2 years ago

<img src="https://tex.z-dn.net/?f=e%20%3D%20mc%20%7B2%7D" id="TexFormula1" title="e = mc {2}" alt="e = mc {2}" align="absmiddle"

serg [7]

Answer:

The equation — E = mc2 — means "energy equals mass times the speed of light squared." It shows that energy (E) and mass (m) are interchangeable.

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2 years ago

Lesson: 1.08Given this function: f(x) = 4 cos(TTX) + 1Find the following and be sure to show work for period, maximum, and minim

Ber [7]

The given function is

$f(x)=4\cos \text{(}\pi x)+1$

The general form of the cosine function is

$y=a\cos (bx+c)+d$

a is the amplitude

2pi/b is the period

c is the phase shift

d is the vertical shift

By comparing the two functions

a = 4

b = pi

c = 0

d = 1

Then its period is

$\begin{gathered} \text{Period}=\frac{2\pi}{\pi} \\ \text{Period}=2 \end{gathered}$

The equation of the midline is

$y_{ml}=\frac{y_{\max }+y_{\min }}{2}$

Since the maximum is at the greatest value of cos, which is 1, then

$\begin{gathered} y_{\max }=4(1)+1 \\ y_{\max }=5 \end{gathered}$

Since the minimum is at the smallest value of cos, which is -1, then

$\begin{gathered} y_{\min }=4(-1)+1 \\ y_{\min }=-4+1 \\ y_{\min }=-3 \end{gathered}$

Then substitute them in the equation of the midline

$\begin{gathered} y_{ml}=\frac{5+(-3)}{2} \\ y_{ml}=\frac{2}{2} \\ y_{ml}=1 \end{gathered}$

The answers are:

Period = 2

Equation of the midline is y = 1

Maximum = 5

Minimum = -3

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2 months ago

The area of the region in the first quadrant between the graph of y = k*sin(x) and x-axis (where k > 0 ) on the closed interv

astraxan [27]

Note that $0\le k\sin x\le k$ in the given closed interval, so the area is exactly given by the definite integral

$\displaystyle\int_0^\pi k\sin x\,\mathrm dx$

(no absolute values needed!)

Integrating gives

$-k\cos x\bigg|_{x=0}^{x=\pi}=-k(\cos\pi-\cos0)=-k(-1-1)=2k$

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2 years ago

Anton [14]

Answer:

The anwser is D.

Step-by-step explanation:

The equation must be greater or equal to eleven but less then or equal to 15.

3 0

2 years ago

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