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telo118 [61]
2 years ago
7

Helppppp : 5×(2-x)+9-7x

Mathematics
2 answers:
lidiya [134]2 years ago
8 0

to solve:

5(2-x)+9-7x: multiply 5 by the 2 and -x.

10-5x+9-7x: combine like terms

the answer is 19-12x

TEA [102]2 years ago
4 0
<span>5×(2-x)+9-7x
=5</span>×2-5<span>×x+9-7x
=10-5x+9-7x
=10+9-(5x+7x)
=19-12x

That's your solution. ^_^</span>
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During taylor's experiment, which piece of data would be qualitative?
Reptile [31]

There isn't any piece of data during Taylor's experiment which can be taken as  qualitative. Thus, correct choice is: Option D: None are qualitative.

<h3>What is qualitative data?</h3>

Qualitative data tells about the quality or characteristic. It is tough to express it numerically or not at all expressible numerically. They are usually catagorical.

In contrast, there is quantitative data which can be expressed numerically.

The problem is missing its option, which are:

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Learn more about qualitative and quantitative data here:

brainly.com/question/12929865

5 0
2 years ago
Please help me I’m stuck!!!
Klio2033 [76]

Answer:

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5 0
2 years ago
A quantity P is an exponential function of time I, such that P = 160 when t = 6 and P = 150 when I = 4. Use the given informatio
Klio2033 [76]

Answer:

  • k = 0.032
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Step-by-step explanation:

Perhaps you want to use the points (t, P) = (4, 150) and (6, 160) to find the parameters P0 and k in the equation ...

  P(t)=P_0\cdot e^{kt}

We know from the given points that we can write the equation as ...

  P(t)=150\left(\dfrac{160}{150}\right)^{(t-4)/(6-4)}=150\left(\dfrac{16}{15}\right)^{\frac{t}{2}-2}\\\\=150\left(\dfrac{16}{15}\right)^{-2}\times\left(\left(\dfrac{16}{15}\right)^{\frac{1}{2}}\right)^t

Comparing this to the desired form, we see that ...

  P_0=150\left(\dfrac{16}{15}\right)^{-2}\approx 131.836\\\\e^{k}=\left(\dfrac{16}{15}\right)^{1/2}\rightarrow k=\dfrac{1}{2}(\ln{16}-\ln{15})\approx 0.0322693

So, the approximate equation for P is ...

  P(t)=131.836\cdote^{0.032t}

And the parameters of interest are ...

  • k = 0.032
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4 0
2 years ago
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