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wolverine [178]
3 years ago
15

How many arches are in an equilateral triangle

Mathematics
2 answers:
defon3 years ago
8 0
There 3 of them
.....................
ELEN [110]3 years ago
3 0
3 arches .............
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If the graph f(x)= 9x^2+37x+41/3x+5 has an oblique asymptote at y=3x+k what is the value of k
ira [324]

Answer:

k=\frac{22}{3}

Step-by-step explanation:

The oblique asymptote of

f(x)=\frac{9x^2+37x+41}{3x+5},

We perform the long division as shown in the attachment.

The quotient is;

3x+\frac{22}{3}

Comparing to 3x+k

Hence the value of k is \frac{22}{3}

6 0
3 years ago
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Find the exponential equation for the given information. Pls help
nikitadnepr [17]

Answer:

y = 12 (2.9)^{x}

Step-by-step explanation:

The standard form of the exponential function is

y = a (b)^{x}

Find a and b by substituting ordered pairs from the table into the equation

Using (0, 12 ), then

12 = ab^{0} ( b^{0} = 1 ), so

a = 12, then

y = 12 b^{x}

Using (1, 34.8 ) , then

34.8 = 12 b^{1} ( divide both sides by 12 )

b = 2.9

Thus exponential function is

y = 12 (2.9)^{x}

3 0
3 years ago
Please help having trouble!
Monica [59]
Long boxes =20 . Divide
8 0
2 years ago
Find the slope for<br> 6y-10x=30
Nuetrik [128]
Slope intercept form is y=5/3x+5
The slope is 5/3
8 0
2 years ago
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A random sample of size 15 taken from a normally distributed population revealed a sample mean of 75 and a sample variance of 25
GREYUIT [131]

Answer:

The upper limit of a 95% confidence interval for the population mean would equal 83.805.

Step-by-step explanation:

The standard deviation is the square root of the variance. Since the variance is 25, the sample's standard deviation is 5.

We have the sample standard deviation, not the population, so we use the t-distribution to solve this question.

T interval:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 15 - 1 = 14

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of 0.95(t_{95}). So we have T = 1.761

The margin of error is:

M = T*s = 1.761*5 = 8.805.

The upper end of the interval is the sample mean added to M. So it is 75 + 8.805 = 83.805.

The upper limit of a 95% confidence interval for the population mean would equal 83.805.

4 0
3 years ago
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