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nlexa [21]
3 years ago
15

“2 less than the quotient of 7 and 3” what’s the expression?

Mathematics
1 answer:
scZoUnD [109]3 years ago
7 0

Answer:

7 divide 3 minus 2

(7 / 3) - 2

The paranthesis aren't really needed

Brainiest plz

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Jenny can swim 10 laps In 5 minutes. How many laps should she swim in 1/3 of an hour?
DIA [1.3K]

1/3 of an hour is 20 minutes. If show 10 laps in 5 minutes, just divide 20 by 5, which gives you 4, then multiply 4 times 10. She can swim 40 laps in 1/3 of an hour. Hope that helps. Mark Brainliest if it does. Thank you!

4 0
3 years ago
Consider r (x) = StartFraction a x Superscript b Baseline + 8 Over c x Superscript d Baseline EndFraction, where a, b, c, and d
madam [21]

Answer:

The correct answer is 0 or choice A on edge

Step-by-step explanation:

right on edge

4 0
3 years ago
The amusement park has 400 employees.seventy-two percent of them are part time employees. How many employees work full-time?
Studentka2010 [4]
400/100=4 so 4 workers are = to 1%
72×4=288 workers are part time
400-288=112 full time workers
6 0
3 years ago
Read 2 more answers
Consider the following three equations:
KengaRu [80]

Answer:

1.

y = 20(11) + 300 = 520

y = 5(11)(11) = 605    

y = 10(1.4)^11 = 404.96

hence the quadratic formula - y = 5x^2 has the largest y-value

2.

y = 20 (14) + 300 = 580

y = 5 (14) (14) = 980

y = 10(1.4)^ 14 = 1111.20   >    this is greater than both equations above

14 is the smallest value of x for which the value of the exponential equation is greater than the values of both the linear and quadratic functions

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
the half-life of strontium-90 is approximately 29 years. how much of a 500 g sample of strontium-90 will remain after 58 years​
Yuliya22 [10]

Answer:  125 g

<u>Step-by-step explanation:</u>

A = P_o\cdot e^{kt}\\\\\text{First, use the given information to find k:}\\\\\bullet A=\dfrac{1}{2}P_o\\\\\bullet k = unknown\\\\\bullet t=29\text{ years}\\\\\dfrac{1}{2}P_o=P_o\cdot e^{k(29)}\\\\\\\dfrac{1}{2}=e^{k(29)}\qquad divided\ both\ sides\ by\ P_o\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=ln\bigg(e^{k(29)}\bigg)\qquad applied\ ln\ to\ both\ sides\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=29k\qquad simplified-ln\ and\ e\ cancel\ out\\\\\\\dfrac{ln\bigg(\dfrac{1}{2}\bigg)}{29}=k\qquad divided\ 29\ from\ both\ sides\\\\\\-0.0239=k

\text{Now, use the following in the equation to solve for A:}\\\\\bullet A=unknown\\\bullet P_o=500\\\bullet k=-0.0239\\\bullet t=58\text{ years}\\\\A=500\cdot e^{(-0.0239)(58)}\\\\.\quad=500\cdot e^{-1.386}\\\\.\quad=125

8 0
3 years ago
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