All categories

3 years ago

What scale factor was applied to the first rectangle to get the resulting image?

Mathematics

2 answers:

arsen [322]3 years ago

6 0

Step-by-step explanation:

6 times x=1.5

x=1/4

x=.25

the first box with a side length of 6 was by .25 to get a side length of 1.5.

Hope that helps :)

Please give brainliestmultiplied

sdas [7]3 years ago

4 0

Answer:

Step-by-step explanation:

In this question, it is given that, A rectangle with a short side of 6. An arrow points to a smaller rectangle with a short side of 1.5 .

The shorter side of first rectangle is of measurement 6 units and of second rectangle, is of size 1.5units .

To find the scale factor, we have to do division of the shorter sides of the first and second rectangle, that is

So the scale factor is 0.25 .

You might be interested in

HELP PLEASEEEEEEEEEEEEE

Colt1911 [192]

Answer:

-5.04199038

Step-by-step explanation:

(y2-y1)/(x2-x1)

Plus, mthway can double check

8 0

3 years ago

In 2-3 sentences describe how you use substitution to solve a system of linear equations?

OverLord2011 [107]

You have to use the substitution method by substituting one Y value with the other

8 0

3 years ago

Suppose that, for a set of data, the range is x and the interquartile range is y. Which statement MUST be true? A) x = y B) x ≤

Vitek1552 [10]

c i suppose the answer

3 0

3 years ago

Read 2 more answers

Two cargo ships spot a signal fire on a small island. The captains know they are 140 feet

sweet [91]

Answer:

add all the angles

Step-by-step explanation:

140+82+78

6 0

3 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$