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2 years ago

PLEASE HELP ME...

Mathematics

2 answers:

adell [148]2 years ago

4 0

Since the reflected figure is moved to the third quardrant, the reflection should be across the x-axis.

Also, if we compare the width and height of the orginial triangle with that of the reflected one, it is clear that it is dilated with a scale factor of 0.25.

erastovalidia [21]2 years ago

4 0

Answer:

Triangle ABC can be mapped onto triangle DEF with a reflection across the x axis followed by a dilation centered at the origin with a scale factor of 0.25 .

Step-by-step explanation:

Given : Graph of two triangle ABC and DEF with coordinates .

To find : Triangle ABC can be mapped onto triangle DEF with a reflection across the______ followed by a dilation centered at the origin with a scale factor of ______.

Solution : We can see from given graph triangle triangle ABC is reflected across x axis.

Scale factor: 0.25

Therefore, Triangle ABC can be mapped onto triangle DEF with a reflection across the x axis followed by a dilation centered at the origin with a scale factor of 0.25 .

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What is the slope of the line that passes through the points

Sidana [21]

Answer:

-5/4

Step-by-step explanation:

8 0

2 years ago

Two ships sail between New York and London. One makes the round trip in 12 days. The other takes 16 days. They both leave London

Mariana [72]

In order to find the answer to this question, you would need to find the Greatest Common Factor of 16 and 12 which would be 4 because 4 is the greatest amount that you can divide both 12 and 16 by. So the soonest both ships will leave together again will be in 4 days.

6 0

3 years ago

Read 2 more answers

A thermometer is taken from a room where the temperature is 24oc to the outdoors, where the temperature is −15oc. After one minu

kap26 [50]

Solution:

Use Newton's Law of Cooling.

T = T_s + (T_0 - T_s)*e^(-kt)

where

T = temperature at any instant

T_s = temperature of surroundings

T_0 = original temperature

t = elapsed time

k = constant

Now, we need to find this constant. We are given that after one hour, the temperature drops to 13° C in a 7°C Environment.

T = 14, T_0 = 24, T_s = -15, t = 1, k = ?

T = T_s + (T_0 - T_s)*e^(-kt)

==> 14 = -15 + (24 - 7)*e^(-k)

==> 14 = 7 + 17*e^(-k)

==> 7 = 17*e^(-k)

==> 7/13 = e^(-k)

==> -k = ln(7/17)

==> k = -ln(7/17) ≈ 0.774

Now,

Let's calculate temperatures!

T = ?, T_0 = 24, T_s = -15, k = 0.773, t = 3

T = T_s + (T_0 - T_s)*e^(-kt)

==> T = -15 + (24 –(-15))*e^[ -(0.774)(2) ]

==> T = -15 + 39*e^(-1.548)

==> T ≈ 15.72° C

This the required answer.

7 0

3 years ago

In 2000, the circulation of a local newspaper was 3,250. In 2001, its circulation was 3,640. In 2002, the circulation was 4,100.

vodka [1.7K]

Percent increase from 2000 to 2001 is 12% and from 2001 to 2002 is 12.6%. Thus period of 2001 to 2002 has higher increase percentage.

Solution:

Given that,

Total circulation of local newspaper in 2000 = 3250

Total circulation of local newspaper in 2001 = 3640

Total circulation of local newspaper in 2002 = 4100

Finding percent of increase in the newspaper’s circulation from 2000 to 2001:

$\text { Percent increase from } 2000 \text { to } 2001=\frac{\text { change in circulation from 2001 - 2000 }}{\text { circulation in } 2000} \times 100$

$\begin{array}{l}{=\frac{3640-3250}{3250} \times 100} \\\\ {=\frac{390}{3250} \times 100=12 \%}\end{array}$

Finding percent of increase in the newspaper’s circulation from 2001 to 2002:

$\begin{array}{l}{\text { Percent increase from } 2001 \text { to } 2002=\frac{\text { change in circulation }}{\text {circulation in } 2001} \times 100} \\\\ {=\frac{4100-3640}{3640} \times 100} \\\\ {=\frac{460}{3640} \times 100=12.6 \%}\end{array}$

As we can see 12% < 12.6%

So period of 2001 to 2002 has higher increase percentage.

7 0

2 years ago

The general solution of 2 y ln(x)y' = (y^2 + 4)/x is

Sav [38]

Replace $y'$ with $\dfrac{\mathrm dy}{\mathrm dx}$ to see that this ODE is separable:

$2y\ln x\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2+4}x\implies\dfrac{2y}{y^2+4}\,\mathrm dy=\dfrac{\mathrm dx}{x\ln x}$

Integrate both sides; on the left, set $u=y^2+4$ so that $\mathrm du=2y\,\mathrm dy$ ; on the right, set $v=\ln x$ so that $\mathrm dv=\dfrac{\mathrm dx}x$ . Then