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lyudmila [28]
3 years ago
10

Solve: log5x^3 - logx^2 = 2 x = 20

Mathematics
2 answers:
Naddika [18.5K]3 years ago
8 0
Yes the answer is 20
Lelu [443]3 years ago
6 0

Answer:

To solve this, we will need to use log properties. We know that subtracting two log equations gives us:

log\frac{5x^{3} }{x^{2} } =2

If log base 10 equals 2, we know that \frac{5x^{3} }{x^{2} } equals 100. We can solve this by:

\frac{5x^{3} }{x^{2} } = 100

5x³=100x²

x³= 20x²

x³-20x²=0

x²(x-20)=0

Thus, the answer is x=20.

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147.50 becuase 8 painting equal 116$ and the smallons equal 31.50 so add them together and you get 147.50

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1. Adam has 2.6 gallons of red paint, 2.35 gallons of white paint, and 2 5/8 gallons of blue paint. If he wants to use up the ca
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A study of recent records of car accidents revealed that average proportion of drivers who were distracted by their phones at th
aksik [14]

Answer:

At least 547 records need to be studied.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

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M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence interval

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In this problem, we have that:

M = 0.04, p = 0.35

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.35*0.65}{n}}

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6 0
3 years ago
a collection of dimes and quarters is worth $19.85. There are 128 coins in all. How many of each type of coin are in the collect
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Number of dimes were 81 and number of quarters were 47

<em><u>Solution:</u></em>

Let "d" be the number of dimes

Let "q" be the number of quarters

We know that,

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

<em><u>Given that There are 128 coins in all</u></em>

number of dimes + number of quarters = 128

d + q = 128 ------ eqn 1

<em><u>Also given that collection of dimes and quarters is worth $19.85</u></em>

number of dimes x value of 1 dime + number of quarters x value of 1 quarter = 19.85

d \times 0.10 + q \times 0.25 = 19.85

0.1d + 0.25q = 19.85  -------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

From eqn 1,

d = 128 - q -------- eqn 3

<em><u>Substitute eqn 3 in eqn 2</u></em>

0.1(128 - q) + 0.25q = 19.85

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12.8 + 0.15q = 19.85

0.15q = 7.05

<h3>q = 47</h3>

Therefore from eqn 3,

d = 128 - q

d = 128 - 47

<h3>d = 81</h3>

Thus number of dimes were 81 and number of quarters were 47

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Answer:

Step-by-step explanation:

Answer is in below attachment

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