Answer:
number of laps = 5/2 = 2.5 laps
Explanation:
First, we will need to convert the minutes to seconds.
5 minutes = 5 * 60 = 300 seconds
Now, we know that she can run 1/3 of a lap in 40 seconds. To know the number of laps that she can run in 300 seconds, all we have to do is cross multiplication as follows:
40 seconds .................> 1/3 laps
300 seconds ................> ?? laps
number of laps = (300 * 1/3) / 40
number of laps = (100) / 40
number of laps = 5/2 = 2.5 laps
Hope this helps :)
The answer is: [C]: -0.7, ⅕, 0.35, ⅔ .
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Explanation:
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<span>
Note that in this correct Answer choice "C" given, we have the following arrangement of numbers:
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</span>→ -0.7, ⅕, 0.35, ⅔ ;
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We are asked to find the "Answer choice" (or, perhaps, "Answer choices?") given that show a set of numbers arranged in order from "least to greatest"; that is, starting with a value that is the smallest number in the arrangement, and sequentially progressing, in order from least to greatest, with the largest (greatest) number in the arrangement appearing as the last number in the arrangement.
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Note the EACH of the 4 (four) answer choices given consists of an arrangement with ONLY one negative number, "- 0.7". Only TWO of the answer choices—Choices "B" and "C"—have an arrangement beginning with the number, "-0.7 "; So we can "rule out" the "Answer choices: [A] and [D]".
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Let us examine: Answer choice: [B]: <span>-0.7, 0.35, ⅕, ⅔ ;
</span>_________________________
Note: The fraction, "⅕" = "2/10"; or, write as: 0.2 .
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The fraction, "⅔" = 0.6666667 (that is 0.6666... repeating; so we often see a "final decimal point" rounded to "7" at some point.
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Through experience, one will be able to automatically look at these 2 (two) fractions and immediately know their "decimal equivalents".
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Otherwise, one can determine the "decimal form" of these values on a calculator by division:
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→ ⅕ = 1/5 = 1 ÷ 5 = 0.2
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→ ⅔ = 2/3 = 2 ÷ 3 = 0.6666666666666667
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For Answer choice: [B], we have:
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→ -0.7, 0.35, ⅕, ⅔ ;
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→ So, we can "rewrite" the arrangement of "Answer choice [B]" as:
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→ -0.7, 0.35, 0.2, 0.666666667 ;
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→ And we can see that "Answer choice: [B]" is INCORRECT; because
"0.2" (that is, "⅕"), is LESS THAN "0.35". So, "0.35" should not come BEFORE "⅕" in the arrangement that applies correctly to the problem.
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Let us examine: Answer choice: [C]: -0.7, ⅕, 0.35, 0.666666667 .
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→ Remember from our previous— and aforementioned—examination of "Answer Choice: [B]" ; that:
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→ ⅕ = 0.2 ; and:
→ ⅔ = 0.666666667
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So, given:
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→ Answer choice: [C]: -0.7, ⅕, 0.35, ⅔ ;
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→ We can "rewrite" this given "arrangement", substituting our known "decimal values for the fractions:
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→ Answer choice: [C]: -0.7, 0.2, 0.35, 0.666666667 ;
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→ As mentioned above, this sequence starts with "-0.7", which is the ONLY negative number in the sequence; as such, the next positive number is correct. Nonetheless, "0.2" (or, "(⅕") is the next number in the sequence, and is greater than "-0.7". The next number is "0.35. "0.35" is greater than "⅕" (or, "0.2"). Then next number is "(⅔)" (or, "0.666666667").
"(⅔)"; (or, "0.666666667") is greater than 0.35.
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This set of numbers: "-0.7, ⅕, 0.35, ⅔" ; is arranged in order from least to greatest; which is "Answer choice: [C]: -0.7, ⅕, 0.35, ⅔" ; the correct answer.
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Answer:
P represents the population in year 0 ⇒ D
Step-by-step explanation:
* Lets explain the exponential growth function
- The exponential growth function is f(x) = a (1 + r)^t, where a is the initial
amount (at t = 0), (1 + r) is the factor of growth , r is the rate of growth
in decimal ant is the time of growth
* Lets solve the problem
∵ The function C(t) = P(1 + r)^t represents the population of
centerville at year t, where P is the initial population and r is the
rate of increase
- Ex: If your investment is increased 10% annually, then that means
each year, your total has multiplied itself by 110% (the growth factor
is 1 + 10/100 = 1.1)
∴ (1 + r) is the factor grows each year
∵ C(t) = P(1 + r)^t
∴ C depends on P(starting population) , r(the increasing rate and
t(the time in year)
∵ r is the rate of increase means the percentage of increasing , then
0 < r < 1
∴ r is not less than 0
∵ P is the initial amount when t = 0
∴ P represents the population in year 0
Answer:
Step-by-step explanation:
Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?
Let p(x)=kpx+dp and q(x)=kqx+dq than
f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7
p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7
(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)
So you want:
−2kpk2q=0
and
kpkq=−1
and
2kpd2p−3kpdq+7=0
Now I amfraid this doesn’t work as −2kpk2q=0 that either kp or kq is zero but than their product can’t be anything but 0 not −1 .
Answer: there are no such linear functions.
