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2 years ago

Find the value of (a)6 (6×5÷3-2) (b) (4÷3)- 3+4

Mathematics

1 answer:

kipiarov [429]2 years ago

5 0

This required PEMDAS

a. 6(30/3 -2) = 6(8) = 48

b. (4/3) - 7 = (4/3 - 21/3) = -17/3

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A,B,D,F
Reasoning, the height should be parallel to the base.

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3 years ago

You need to get to class, 200 meters away, and you can only walk in the hallways at about 1.5 m/s. (if you run any faster, you’l

RUDIKE [14]

Answer:

Time = 133.3 secs

Step-by-step explanation:

Speed = Distance / Time

Rearranging the formula

Time = Distance / Speed

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Time = 133.3 secs

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3 years ago

Read 2 more answers

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Step-by-step explanation:

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2 years ago

A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, wri

GuDViN [60]

Answer:

1. $\frac{dy}{dt}=ky$

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

$\frac{dy}{dt}$ =Number of bacteria per unit time

$\frac{dy}{dt}\proportional y$

$\frac{dy}{dt}=ky$

Where k=Proportionality constant

2. $\frac{dy}{y}=kdt$ ,y'(0)=100

Integrating on both sides then, we get

$lny=kt+C$

We have y(0)=200

Substitute the values then , we get

$ln 200=k(0)+C$

$C=ln 200$

Substitute the value of C then we get

$ln y=kt+ln 200$

$ln y-ln200=kt$

$ln\frac{y}{200}=kt$

$\frac{y}{200}=e^{kt}$

$y=200e^{kt}$

Differentiate w.r.t

$y'=200ke^{kt}$

Substitute the given condition then, we get

$100=200ke^{0}=200 \;because \;e^0=1$

$k=\frac{100}{200}=\frac{1}{2}$

$y=200e^{\frac{t}{2}}$

Substitute t=2

Then, we get $y=200e^{\frac{2}{2}}=200e$

$y=200(2.718)=543.6=543.6$

e=2.718

Hence, the number of bacteria after 2 hours=543.6

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3 years ago

in klm the measure of m=90 the measure of l=68 and mk=2.5 feet. find the length of kl to the nearest tenth of a foot

frutty [35]

Answer:

2.7

Step-by-step explanation:

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2 years ago

Find the value of (a)6 (6×5÷3-2) (b) (4÷3)- 3+4​

Find the value of (a)6 (6×5÷3-2) (b) (4÷3)- 3+4