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2 years ago

Is the product of - 7 and 4 less than - 7,between 7 and 4, or greater than 4?

Mathematics

2 answers:

Tcecarenko [31]2 years ago

3 0

Answer:

less than -7

Step-by-step explanation:

The product of -7 and 4 is -28, therefore it is less than -7.

zlopas [31]2 years ago

3 0

Answer:

less than -7

Step-by-step explanation:

-7 times 4 = -28

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Successfulness of the Competition policy in South Africa- support and graphs

jenyasd209 [6]

Answer:

Answer to the following question is a follows;

Step-by-step explanation:

The following are a few examples of how South Africa's competitiveness policy has been successful:

⇒ Consumers or buyers were given a variety of product options as well as competitive prices.

⇒ In 1984, practises like horizontal cooperation and resale price maintenance and control were ruled illegal.

6 0

2 years ago

Express 35.4% as a decimal.

lakkis [162]

It is because the 35.4% divided the 100% total is B. 0.354

8 0

3 years ago

Read 2 more answers

$10,000 for 20 years at 5% compounded annually

saveliy_v [14]

Answer:

$26532.98

Step-by-step explanation:

<u>Given:</u>

Principal = $10000
Profit rate = 5% PA compounded
Time = 20 years
Compounds = 20*1 = 20

10000*(1 + 5/100)²⁰ ≈ 26532.98

3 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago

a student answers 75% of the questions on a math exam correctly. if he answers 30 questions correctly. how many questions are on

Komok [63]

There is 40 questions on the exam

8 0

2 years ago