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andre [41]
3 years ago
12

20−6+4k=2−2k plz solve

Mathematics
2 answers:
Orlov [11]3 years ago
8 0

Answer:

-2

Step-by-step explanation:

jekas [21]3 years ago
4 0

Answer:

k=-2

Step-by-step explanation:

isolate the viarable by dividing each side by factors that  contain the  variable

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Both of them are the first answer choice
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Log6 8 + log6 27 as single logarithm
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Log[8/(36*3)] Log[2/27] Hope I really help you !
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The midpoint of UV is point W. What are the coordinates of point V?
madreJ [45]

Since you don't provide the coordinates of the point W, I will help you in a general form anyway. In the Figure below is represented the segment that matches this problem. We have two endpoints U and V. So, by using the midpoint formula we may solve this problem:

Midpoint=W=W(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})=W(x_{3}, y_{3}) \\ \\ where:\\ \ U=U(x_{1}, y_{1}) \ and \ V=V(x_{2}, y_{2}) \\ \\ and: \\ x_{3}=\frac{x_{1}+x_{2}}{2} \\ y_{3}=\frac{y_{1}+y_{2}}{2}

Therefore:

x_{2}=2x_{3}-x_{1} \\ y_{2}=2y_{3}-y_{1}

So we know x_{3} \ and \ y_{3} but we also must know x_{1} \ and \ y_{1}

Finally, knowing the points U and W we can find the endpoint V.

7 0
3 years ago
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A restaurant has 9 pounds of onion. The restaurant's manager ordered 4 more pounds of onion. The onions then were mixed and kept
Iteru [2.4K]

Answer:

6.5

Step-by-step explanation:

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4 0
3 years ago
A researcher determines that students are active about 60 + 12 (M + SD) minutes per day. Assuming these data are normally distri
rjkz [21]

Answer:

The correct option is (b).

Step-by-step explanation:

If X \sim N (µ, σ²), then Z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z \sim N (0, 1).

The distribution of these z-variate is known as the standard normal distribution.

The mean and standard deviation of the active minutes of students is:

<em>μ</em> = 60 minutes

<em>σ </em> = 12 minutes

Compute the <em>z</em>-score for the student being active 48 minutes as follows:

Z=\frac{X-\mu}{\sigma}=\frac{48-60}{12}=\frac{-12}{12}=-1.0

Thus, the <em>z</em>-score for the student being active 48 minutes is -1.0.

The correct option is (b).

4 0
3 years ago
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