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3 years ago

Write each exponential expression as a decimal.

Mathematics

1 answer:

HACTEHA [7]3 years ago

8 0

1. 100 2. 1000 3. 100000 4. 10000000 here is the first part

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Which expression is equivalent to 2(3x − 1.75) − 3(1.5x − 2.5)?

lara [203]

Answer:

1.5x+4

Step-by-step explanation:

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3 years ago

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I forgot how to solve these. Please help.

AnnyKZ [126]

This is the answer for both number 9 and number 11.
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3 years ago

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A bacteria colony increases in size at a rate of 4.0581e1.6t bacteria per hour. If the initial population is 36 bacteria, find t

Harrizon [31]

Answer:

1560

Step-by-step explanation:

The rate of Increase of the population (P) of the bacteria is given as:

$\frac{dP}{dt} =4.058e^{1.6t}$

$dP =4.058e^{1.6t}}dt\\Taking\: Integrals\\\int dP =4.058 \int e^{1.6t}dt\\P(t)=\frac{4.058}{1.6} (e^{1.6t} +K)\\P(t)=2.53625 (e^{1.6t} +K)$

Where k is a constant of Integration.

At t=0, P(t)=36

$36=2.53625 (e^{1.6*0} +K)\\36=2.53625 (e^{0} +K)\\36=2.53625 (1 +K)\\36=2.53625 +2.53625K\\K=13.19$

Therefore:

$P(t)=2.53625 (e^{1.6t} +13.19)\\At \: t=4\\P(4)=2.53625 (e^{1.6*4} +13.19)\\=1559.88\\P(4)=1560$

3 0

3 years ago

All of the expressions that have the same value as LaTeX: 892\div8892 ÷ 8. Group of answer choices LaTeX: 8920\div808920 ÷ 80 La

algol13

Answer:

$(a)$ $8920\div80$

$(e)$ $0.892\div0.008$

Step-by-step explanation:

Given

$892\div8$

Required

Select equivalent expression

$8920\div80$

$894\div10$

$89.2\div0.08$

$8.92\div0.8$

$0.892\div0.008$

Start by solving the given expression (using a calculator)

$892\div8$ $= 111.5$

Then we solve the given options (using a calculator)

$(a)$ $8920\div80$

$8920\div80$ $= 111.5$

$(b)$ $894\div10$

$894\div10$ $= 89.4$

$(c)$ $89.2\div0.08$

$89.2\div0.08$ $= 1115$

$(d)$ $8.92\div0.8$

$8.92\div0.8$ $= 11.15$

$(e)$ $0.892\div0.008$

$0.892\div0.008$ $= 111.5$

The equivalent expressions are (a) and (e)

5 0

2 years ago

Plz help LOL <br> 5(x-3)-x=

V125BC [204]

5(x-3)-x =
Multiply Everything In The Parentheses By 5
5x-15-x

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3 years ago

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