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3 years ago

Please help ...........

Mathematics

1 answer:

Ludmilka [50]3 years ago

5 0

Answer:

Step-by-step explanation:

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The product of 3x and (3x+1) is a t least 35

Leokris [45]

Product means the sum of two things i.e. the product of 1 and 2 is 3 <span>(1 + 2 = 3)

So the product of 3x and (3x+1) can be written as:
35 < 3x + 3x + 1
^ Greater than 35 i.e. at least 35
35 < 6x + 1
34 < 6x
34/6 < x
17/3 < x
5.67 (2dp) < x

</span>

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4 years ago

Kendra bought 235 donuts for $120. Her profit was $35 once she had sold 100 donuts. Which equation below represents Kendra’s pro

marysya [2.9K]

The last one
P(n) = 1.55n-120

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3 years ago

Help with this please question 3

MrRissso [65]

3a. Jackson's gross pay for 40 hours is
(40 h)×($12.20/h) = $488.00

3b. An equation for Jackson's pay could be
372.10 = h×12.20
This can be solved for h by dividing by 12.20
372.10/12.20 = h = 30.5
Jackson worked 30.5 hours last week.

3c. The total of hours shown is
(8) + (7 1/4) + (8 1/2) = (8 +7 +8) +(0.25 +0.50) = 23.75
The number of hours missing is the difference between this number and the hours found in part (b).
30.50 - 23.75 = 6.75
There are 6 3/4 hours missing from Jackson's time sheet for last week. This is less than a full day's work, so seems reasonable.

_____
Of course, you know the decimal - fraction conversions:
1/4 = 0.25
1/2 = 0.50
3/4 = 0.75

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3 years ago

An isosceles triangle has perimeter T and sides of length x, x, and y. Which of the following represents x in terms of T and y?

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C. T-2y/2 easy peazy lemon squeezey

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3 years ago

A breeder reactor converts uranium-238 into an isotope of plutonium-239 at a rate proportional to the amount of uranium-238 pres

topjm [15]

Let $U(t)$ denote the amount of uranium-238 in the reactor at time $t$ . As conversion to plutonium-239 occurs, the amount of uranium will decrease, so the conversion rate is negative. Because the rate is proportional to the current amount of uranium, we have

$\dfrac{\mathrm dU}{\mathrm dt}=-kU$

where $k>0$ is constant. Separating variables and integrating both sides gives

$\dfrac{\mathrm dU}U=-k\,\mathrm dt\implies\ln|U|=-kt+C\implies U=Ce^{-kt}$

Suppose we start some amount $u$ . This means that at time $t=0$ we have $U(0)=u$ , so that

$u=Ce^{-0k}\implies C=u\implies U=ue^{-kt}$

We're given that after 10 years, 99.97% of the original amount of uranium remains. This means (if $t$ is taken to be in years) for some starting amount $u$ ,

$0.9997u=ue^{-10k}\implies k=-\dfrac{\ln(0.9997)}{10}$

The half-life is the time $t_{1/2}$ it takes for the starting amount $u$ to decay to half, $0.5u$ :

$0.5u=ue^{-kt_{1/2}}\implies t_{1/2}=-\dfrac{\ln(0.5)}k=\dfrac{10\ln2}{\ln(0.9997)}$

or about 23,101 years. Notice that it doesn't matter what the actual starting amount is, the half-life is independent of that.

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3 years ago