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matrenka [14]
3 years ago
11

What is the value of eniging

Mathematics
1 answer:
rjkz [21]3 years ago
4 0
Eniging?????????????????
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Match each slope and y-intercept with the appropriate linear equation​
aleksandr82 [10.1K]

Linear equation: y = mx + c

m = slope

c = y-intercept

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How would logx (49) = x look like in exponential form
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I'm assuming the given log equation is \log_{\text{x}}(49) = \text{x}

If so, then the exponential form is \text{x}^{\text{x}} = 49

This is because the general form \log_{\text{b}}(\text{x}) = \text{y} transforms into \text{y} = \text{b}^{\text{x}}

For both equations, the 'b' is the base.

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A school has 510 male students and 850 female students. If a student from the school is selected at random, what is the probabil
Colt1911 [192]

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0.375

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510 male out of a total of 1360

510/1360 = 0.375

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Manuel created a factor tree and wrote the prime factorization of 60 shown below.
bagirrra123 [75]

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The second one or B- "He did not find the prime factors of 4."

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I got it right on Edge2020

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The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct
lukranit [14]

Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

y=\dfrac{x}{1+x}

Point P\left(1,\dfrac{1}{2}\right) lies on the curve.

Point Q\left(x,\dfrac{x}{1+x}\right) is an arbitrary point on the curve.

The slope of the secant line PQ is

\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]

1. If x=0.5, then the slope is

\dfrac{1}{2(0.5+1)}=\dfrac{1}{3}\approx 0.3333

2. If x=0.9, then the slope is

\dfrac{1}{2(0.9+1)}=\dfrac{1}{3.8}\approx 0.2632

3. If x=0.99, then the slope is

\dfrac{1}{2(0.99+1)}=\dfrac{1}{3.98}\approx 0.2513

4. If x=0.999, then the slope is

\dfrac{1}{2(0.999+1)}=\dfrac{1}{3.998}\approx 0.2501

5. If x=1.5, then the slope is

\dfrac{1}{2(1.5+1)}=\dfrac{1}{5}\approx 0.2

6. If x=1.1, then the slope is

\dfrac{1}{2(1.1+1)}=\dfrac{1}{4.2}\approx 0.2381

7. If x=1.01, then the slope is

\dfrac{1}{2(1.01+1)}=\dfrac{1}{4.02}\approx 0.2488

8. If x=1.001, then the slope is

\dfrac{1}{2(1.001+1)}=\dfrac{1}{4.002}\approx 0.2499

7 0
3 years ago
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