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Ivan
4 years ago
12

Write an equation for the line that is perpendicular to the given line and that passes through the given point. y = 2x + 7; (0,

0)
Mathematics
1 answer:
Scilla [17]4 years ago
8 0

Answer:

y = 1/-2x + 0 or y = 1/-2x

Step-by-step explanation:

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Initially, there were only 86 weeds in the garden. The weeds grew at a rate of 29% each week. The following function represents
Orlov [11]

Step-by-step answer:

The base of the exponential function is 1.29 for 7 days, as in

f(x) = 86*(1.29)^x

The new rate for days can be calculated by dividing x by 7 (where x remains the number of weeks), namely

f(x) = 86*1.29^(x/7)

Using the law of exponents, b^(x/a) = b^(x*(1/a)) = (b^(1/a))^x

we simplify by putting b=1.29, a=7 to get

f(x) = 86*(1.29^(1/7))^x

f(x) = 86*(1.037)^x      since 1.29^(1/7) evaluates to 1.037

Rounding 1.037 to 1.04 we get a (VERY) approximate function

f(x) = 86 * (1.04^x)

1.04 is very approximate because 1.04^7 is supposed to get back 1.29, but it is actually 1.316, while 1.037^7 gives 1.2896, much closer to 1.29.

7 0
3 years ago
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Anyone have a real answer for this??
Otrada [13]
I am not sure but I believe it should be close to 100%. I don´t really know

4 0
4 years ago
How is the product of 3 and 2 shown on a number line?
Bas_tet [7]

Answer:

A

Step-by-step explanation:

3 0
4 years ago
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What is the answer to the math problem 5+9(2+4)?
IceJOKER [234]

Answer:

59

Step-by-step explanation:

Simplify the following:

5 + 9 (2 + 4)

2 + 4 = 6:

5 + 9×6

9×6 = 54:

5 + 54

5 + 54 = 59:

Answer:  59

6 0
3 years ago
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Expanding logarithmic Expression In Exercise,Use the properties of logarithms to rewrite the expression as a sum,difference,or m
aliina [53]

Answer:

\ln x(x^2 + 1)^\frac{1}{3} = \ln x + \dfrac{1}{3} \ln (x^2 + 1)

Step-by-step explanation:

We are given the following expression in the question

\ln x(x^2 + 1)^\frac{1}{3}

Logarithmic Properties:

\log (ab) = \log a + \log b\\\\\log \dfrac{a}{b} = \log a - \log b\\\\\log (a^b) = b\log a

We have to simplify the given expression

\ln x(x^2 + 1)^\frac{1}{3}\\=\ln x + \ln (x^2 + 1)^\frac{1}{3}\\\ln x + \dfrac{1}{3} \ln (x^2 + 1)

\ln x(x^2 + 1)^\frac{1}{3} = \ln x + \dfrac{1}{3} \ln (x^2 + 1)

6 0
3 years ago
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