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3 years ago

Pls help me

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

7 0

Answer:

area = 2604 in²

Step-by-step explanation:

area = L x W

area = 42 x 62 = 2604 in²

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Which point on the numbee line represents the product (-4)(-2)(-1)

erica [24]

Answer:-4*-2=+8*-1=-8

cevap A

Step-by-step explanation:A

4 0

3 years ago

Write a real-world problem to compare fractions.then solve the problem

andrew11 [14]

The fraction computed shows that the total quantity of meat that Joy bought is 5/6kg.

<h3>How to solve the fraction?</h3>

The fraction can be explained thus: Joy bought 1/2kg of meat from a store and bought 1/3kg of meat from store B. What is the total quantity of meat that Joy bought?

The total quantity of meat that Joy bought will be:

= 1/2 + 1/3

= 5/6

In conclusion, the correct option is 5/6kg.

Learn more about fractions on:

brainly.com/question/78672

6 0

2 years ago

I’m kinda confused. Can someone help.

horsena [70]

Answer:

PQ = $PM\cos MPQ + MQ\cos MQP=2PM\cos MQP$

Step-by-step explanation:

given MQR = 60, PQR = 75

MQP = 75 -60= 15 $QR = ((18(1+\cos 30))^{2}-(2\cos 15)^{2})^{\frac{1}{2}}$

WE KNOW THAT MQR +QRM +QMR = 180

MPQ = MQP AS ANGLE OPPOSITE TO EQUAL SIDES ARE EQUAL

THEREFORE QMR = 90 - 60 = 30

therefore PQM = 75 - 60 = 15

PM = PQ because M is the mid point

therefore PR = PM + $MQ\cos QMR$

PR = $18(1+\cos 30)$

$QR = (PR^{2}-PQ^{2}))^{\frac{1}{2}}$

PQ = $PM\cos MPQ + MQ\cos MQP=2PM\cos MQP$

5 0

3 years ago

A rumor spreads through a small town. Let y(t) be the fraction of the population that has heard the rumor at time t and assume t

sladkih [1.3K]

Answer:

The answer is shown below

Step-by-step explanation:

Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

$\frac{dy}{dt}\ \alpha\ y(1-y)$

$\frac{dy}{dt}=ky(1-y)$

where k is the constant of proportionality, dy/dt = rate at which the rumor spreads

$\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}} =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\$