How do you find the perimeter and area?

Answer:

RTP: 1 + cos2x = 2/1+tan^2x

LHS  =  1 + cos2x

        =  1 + 2cos^2x - 1

        =  2cos^2x

RHS = 2/1+tan^2x

        = 2/sec^2x

        = 2 x cos^2x

        = 2cos^2x

2cos^2x=2cos^2x

LHS = RHS

Step-by-step explanation:

Answer:

{(-5, 5), (0, 5), (5, 5)}  Step-by-step explanation:

A trend line shows the general pattern of the data, but does not try to connect all the data points.

The correct statement about trend lines is:

A. A trend line shows the general pattern of the data, but does not try to connect all the data points.

A trend line is used to represent the overall trend or pattern in a set of data. It is not necessary for all the data points to fall exactly on the trend line. The trend line provides a visual representation of the relationship between the variables being measured, such as the change over time.

To write a quadratic function in standard form with zeros 6 and -8, we use the fact that the zeros of a quadratic function are the values of x for which the function equals zero. The quadratic function in standard form with zeros 6 and -8 is x^2 + 2x - 48 = 0.

To write a quadratic function in standard form with zeros 6 and -8, we use the fact that the zeros of a quadratic function are the values of x for which the function equals zero. The standard form of a quadratic function is ax^2 + bx + c = 0. Since the zeros are 6 and -8, the factors of the quadratic function will be (x - 6) and (x + 8). Multiplying these factors gives us (x - 6)(x + 8) = 0. Expanding and simplifying this expression, we get x^2 + 2x - 48 = 0. Therefore, the quadratic function in standard form with zeros 6 and -8 is x^2 + 2x - 48 = 0.

The transaction most likely to bring in the most income is from individuals who use credit cards offering air miles and charge over $2,000 per month.

Based on the data provided, the type of transaction likely to bring in the most income during a month involves credit cards that offer a mile of air travel for every dollar charged, particularly among users who spend over $2,000 monthly. To assess this, we can perform a simple calculation. With 30% of respondents charging more than $2,000, and 80% of those using the air miles card, we can infer that 0.30 × 0.80 = 0.24 or 24% of all respondents represent this high-earning transaction category.

Gathering data at the local supermarket on the total amounts of grocery receipts or from cashiers could provide additional insights into shopping trends and payment methods. The student can collect the data, analyze spending habits, and cross-reference those with preferred payment methods to identify the most profitable transaction. However, the information about the income related to U.S. international financial flows, like exports in goods and services or income received by U.S. investors, while related to economic transactions, does not directly answer the question about the most income-producing type of credit card usage during a month.

The most income-generating transaction per month is likely to be credit card transactions offering air miles for users who charge more than $2,000.

Analysis of the Most Income-Generating Transaction

According to the provided data, the transaction likely to bring in the most income during a month would be transactions made with a credit card that provides a mile of air travel for every dollar charged, especially for users who charge more than $2,000 per month. The data indicates that 30% of respondents charge more than $2,000 monthly, and of those, a significant 80% use a credit card offering air miles. Given this high rate, it is reasonable to infer that a credit card transaction yielding air miles for users spending over $2,000 a month would generate the most income.

To illustrate the calculation: If we consider 100 respondents, 30 of them would charge over $2,000, and 80% of these 30 people use the air miles card, resulting in 24 people. If each of these 24 people spends exactly $2,000, this equates to $48,000 possibly earned.

By applying logarithmic properties and solving a resulting quadratic equation, we find the solutions x = 3  and x = -7 for the given logarithmic equation.

To solve the logarithmic equation [tex]\( \log_4(x+5) + \log_4(x-1) = 2 \),[/tex] we can use the properties of logarithms to condense the equation into a single logarithm, then solve for x.

First, we apply the product rule of logarithms, which states that [tex]\( \log_a(b) + \log_a(c) = \log_a(bc) \)[/tex], to condense the two logarithms on the left side:

[tex]\[ \log_4((x+5)(x-1)) = 2 \][/tex]

Now, we have:

[tex]\[ (x+5)(x-1) = 4^2 \][/tex]

[tex]\[ (x+5)(x-1) = 16 \][/tex]

Next, we expand the left side of the equation:

[tex]\[ x^2 - x + 5x - 5 = 16 \][/tex]

[tex]\[ x^2 + 4x - 5 - 16 = 0 \][/tex]

[tex]\[ x^2 + 4x - 21 = 0 \][/tex]

Now, we can solve this quadratic equation for x. We can use the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

For [tex]\( x^2 + 4x - 21 = 0 \), \( a = 1 \), \( b = 4 \), and \( c = -21 \).[/tex]

Plugging in these values, we get:

[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-21)}}{2(1)} \][/tex]

[tex]\[ x = \frac{-4 \pm \sqrt{16 + 84}}{2} \][/tex]

[tex]\[ x = \frac{-4 \pm \sqrt{100}}{2} \][/tex]

[tex]\[ x = \frac{-4 \pm 10}{2} \][/tex]

[tex]\[ x = \frac{6}{2} \][/tex] or [tex]\( x = \frac{-14}{2}[/tex]

x = 3  or x = -7

So, the solutions to the equation are  x = 3  and x = -7 .

The question probable may be:

Given a log equation solve it:

[tex]\( \log_4(x+5) + \log_4(x-1) = 2 \),[/tex]

Step-by-step explanation:

Equation of straight line is given by y = mx + c

    Slope of line , m = -1/5

    y- intercept, c = 5

Substituting

                      [tex]y=mx+c\\\\y=\frac{-1}{5}x+5\\\\y=\frac{-x+25}{5}\\\\5y=-x+25\\\\x+5y-25=0[/tex]

x + 5 y - 25 = 0

Option C is the correct answer.

The correct answer is: 8

Explanation:

Given mixed number:

[tex] 7\frac{4}{5} [/tex]

In order to convert a mixed number into a decimal number, follow the steps mentioned below:

Multiple 7 with 5 and add the numerator to the answer you get by multiplying 7 with 5. After that, write that result in the numerator and divide it with 5:

Step-1:

Multiply 7 with 5.

7*5 = 35

Step-2:

Add numerator value, which is 4 in this case, in 35

35 + 4 = 39

Step-3:

Rewrite the fraction.

[tex] \frac{39}{5} [/tex]

Step-4:

Calculate.

[tex] \frac{39}{5} = 7.8 [/tex]

Now, as the number at the tenth place after the decimal point is greater than 5, we have to round it off to 8.0.

Hence the correct answer is 8.

Rounding 7 4/5 to the nearest whole number gives 8.

To round the fraction 7 4/5 to the nearest whole number, we examine the fractional part (4/5) of the mixed number.

The fractional part, 4/5, is greater than or equal to 1/2. In such cases, when the fractional part is greater than or equal to 1/2, we round the whole number up to the next higher whole number.

In this case, the whole number part is 7. Since the fractional part is greater than 1/2, we round the whole number up to the next higher whole number, which is 8.

Therefore, rounding 7 4/5 to the nearest whole number gives us 8.

In summary, when rounding a mixed number to the nearest whole number, we look at the fractional part. If it is greater than or equal to 1/2, we round the whole number up. If it is less than 1/2, we round the whole number down. In this case, since 4/5 is greater than 1/2, we round 7 up to 8.

To learn more about fractions;

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To find m∠1, you typically use geometry knowledge, including understanding the properties of polygons and triangles. Without a specific diagram or context, however, it's impossible to provide explicit steps or an exact answer for m∠1.

Without specific figures, it's impossible to determine exact angle measures. However, in general to find m∠1, you will need to draw upon your knowledge of geometry, particularly angle properties. If the angle is part of a polygon, its measure can often be determined by the other angle measures in the shape. Similarly, if the angle is part of a triangle, you could use the fact that the sum of all angles in a triangle is 180°. If angle properties like supplementary or complementary are mentioned, it means two angles add up to 180° or 90° respectively. Without a diagram or further context, I cannot provide particular steps to find m∠1.

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___________________

Step-by-step explanation:

|3x – 5| > 10 ⇔ 3x – 5 > 10 or 3x – 5 < -10.

:)

Answer: 3x-5<-10 or 3x-5>10

Step-by-step explanation:

Answer:

17, 14

Step-by-step explanation:

Answer:

40%

Step-by-step explanation:

formula change/original

change: 10

original: 25

10/25 = 0.4

0.4 = 40%

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Answer:  Option '4' is correct.

Step-by-step explanation:

Since we have given that

Total number of games =50

Firstly, he played 20 games.

Percentage of games they won = 60%

Number of games they won is given by

[tex]\dfrac{60}{100}\times 20\\\\=0.6\times 20\\\\=12[/tex]

Remaining number of games = 50-20=30

Percentage of remaining games they won = 80%

So, number of games they won is given by

[tex]\dfrac{80}{100}\times 30\\\\=0.8\times 30\\\\=24[/tex]

So, total number of games they won is given by

12+24 = 36

So, Percentage of games they won for the entire season will be

[tex]\dfrac{36}{50}\times 100\\\\=36\times 2\\\\=72\%[/tex]

Hence, Option '4' is correct.

The student's question is about simplifying a complex algebraic expression that involves exponents and basic arithmetic operations. The approach to simplification entails applying the rules of exponents and arithmetic operations systematically.

The student's question appears to involve simplifying an algebraic expression with exponents and performing operations such as division and multiplication. While the question contains parts that are not directly relevant to the simplification process, we can still address the general approach to simplify an expression like a^4 divided by 2a^2 and then multiplied by 4a^4 times 8a^3, plus an additional term 1/2a plus 4a^2.

To simplify such an expression, one would apply the rules of exponents and arithmetic operations step by step. For instance, when dividing exponents with the same base, you subtract the exponents, and when multiplying, you add the exponents. This process typically results in a much simpler expression that can easily be understood or further manipulated.