What is the solution set of the following equation?

4/5x^2=2x-4/5

1. {1, 2}
2. {1, 1/2}
3. {2, 1\2}

Shane and his family spent a total of $103 at the movie theatre, which includes the cost of 6 tickets at $12 each and $31 on popcorn and drinks.

Shane and his family bought 6 tickets for $12 each, thus the total cost of the tickets is 6 * $12 which equals $72. Afterwards, they spent an additional $31 on popcorn and drinks. To find out how much they spent altogether at the movie theatre, we have to add the cost of the tickets and the cost of the popcorn and drinks. Therefore, $72 (for the tickets) + $31 (for the snacks) = $103. So, Shane and his family spent a total of $103 at the movie theatre.

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A percentage is a way to describe a part of a whole. The best option for the customer to buy is option A.

A percentage is a way to describe a part of a whole. such as the fraction ¼ can be described as 0.25 which is equal to 25%.

Let the cost price of a cereal box be represented by x.

Given that the box usually contains 450 grams of cereal. Therefore, the unit rate for each option can be written as,

Option A

Weight of cereal = 20% more cereal

= 450 grams + 20% of 450 grams

= 450 grams + 90 grams

= 540 grams

Unit rate for option A = $x / 540 grams

Option B

Cost price of cereal = 15% off the price

= x - 15% of x

= x + 0.15x

= 0.85x

Unit rate for option A = $0.85x / 450 grams

Assume that option A is good. Now, if we subtract unit rate of option B from unit rate of option A then two of the cases are possible, if the value of the result is positive then the unit rate of option A is bigger while if the value is negative unit rate of option B is bigger.

Difference = (x/540) - (0.85x/450)

= (450x - 459x)/243000

= -9x/243,000

= -x/27,000

Further, since the difference is negative, therefore, the unit rate of option B is bigger. But since the value of option B is bigger it is costly and not the correct option to buy.

Hence, the best option for the customer to buy is option A.

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Answer:

WrItE wHaT yOu FeElLlLlL

Step-by-step explanation:

Answer:

K′(−6, −4); L′(−3, −3)

Step-by-step explanation:

Got it right on the test

we have in this problem

[tex] 3x+1 > 10 [/tex] ------> inequality [tex] 1 [/tex]

[tex] 4x \leq 20 [/tex] ------> inequality [tex] 2 [/tex]

Solve the inequality [tex] 1 [/tex]

[tex] 3x+1 > 10\\ 3x > 10-1\\ 3x > 9\\ x > 3 [/tex]

the solution is the interval------> (3,∞)

Solve the inequality [tex] 2 [/tex]

[tex] 4x \leq 20\\\\ x \leq \frac{20}{4} \\\\ x \leq 5 [/tex]

the solution is the interval------> (-∞,5]

Find the solution for the system

(3,∞) ∩ (-∞,5]--------> (3,5]

therefore

the answer is the option

A. A number line with an open circle on 3, a closed circle on 5, and shading in between.

The solution to the equation 12^x - 2 = 20, when solved for x and rounded to the nearest ten-thousandth, is x = 1.4624.

To solve the equation 12^x - 2 = 20, we first need to isolate the exponential term. This can be done by adding 2 to both sides of the equation resulting in 12^x = 22. Next, we take the logarithm of both sides to solve for x. We can use any logarithmic base, but common practice is to use base 10 or the natural logarithm (base e). The next steps are:

Calculating the value on a calculator gives us x ≈ 1.4624. Therefore, the solution to the equation, rounded to the nearest ten-thousandth, is x = 1.4624.

The correct answer is option (b) "x ≈ 1.2056."

To solve the equation [tex]\(12^{(x-2)} = 20\)[/tex], we can take the logarithm of both sides.

Let's use the natural logarithm (ln) for this:

[tex]\[ \ln(12^{(x-2)}) = \ln(20) \][/tex]

Using the logarithm property [tex]\( \log_a(b^c) = c \cdot \log_a(b) \)[/tex], we can rewrite the left side:

[tex]\[ (x-2) \cdot \ln(12) = \ln(20) \][/tex]

Now, solve for (x):

[tex]\begin{aligned}& x-2=\frac{\ln (20)}{\ln (12)} \\& x=\frac{\ln (20)}{\ln (12)}+2\end{aligned}[/tex]

Now, calculate the value:

[tex]\begin{aligned}& x \approx \frac{\ln (20)}{\ln (12)}+2 \\& x \approx \frac{2.9957}{2.4849}+2 \\& x \approx 1.2056\end{aligned}[/tex]

So, the solution to the equation is approximately [tex]\( x \approx 1.2056 \)[/tex]

Complete Question:

Solve [tex]\(12^{(x-2)} = 20\)[/tex]. Round to the nearest ten-thousandth.

a. x ≈ 1.0792

b. x ≈ 1.2056

c. x ≈ 3.2056

d. x ≈ 1.3010

there are [tex]\( \frac{10}{3} \) or \( 3 \frac{1}{3} \) servings of \( \frac{1}{4} \)-cup in \( \frac{5}{6} \)[/tex] cup of ice cream.

To find out how many [tex]\( \frac{1}{4} \)-cup servings are in \( \frac{5}{6} \)[/tex] cup of ice cream, we need to divide [tex]\( \frac{5}{6} \) by \( \frac{1}{4} \)[/tex].

We can do this by converting [tex]\( \frac{5}{6} \)[/tex] to have the same denominator as [tex]\( \frac{1}{4} \)[/tex], which is 6, and then dividing:

[tex]\[ \frac{5}{6} \div \frac{1}{4} \][/tex]

To make the denominators the same, we multiply both the numerator and denominator of [tex]\( \frac{5}{6} \) by \( 4 \):[/tex]

[tex]\[ \frac{5 \times 4}{6 \times 4} \div \frac{1}{4} \][/tex]

[tex]\[ = \frac{20}{24} \div \frac{1}{4} \][/tex]

Now, we can divide [tex]\( \frac{20}{24} \) by \( \frac{1}{4} \):[/tex]

[tex]\[ \frac{20}{24} \div \frac{1}{4} \][/tex]

[tex]\[ = \frac{20}{24} \times \frac{4}{1} \][/tex]

[tex]\[ = \frac{20 \times 4}{24 \times 1} \][/tex]

[tex]\[ = \frac{80}{24} \][/tex]

Now, we simplify the fraction:

[tex]\[ \frac{80}{24} = \frac{10 \times 8}{12 \times 2} \][/tex]

[tex]\[ = \frac{10}{3} \][/tex]

Therefore, there are [tex]\( \frac{10}{3} \) or \( 3 \frac{1}{3} \) servings of \( \frac{1}{4} \)-cup in \( \frac{5}{6} \)[/tex] cup of ice cream.

Answer:

x = 21

Step-by-step explanation:

If RECT is a rectangle, each of its interior angles are right angles, 90°. Since ∠RTC is split into two angles, the sum of the two angles must be 90°. (Angles that add to 90° are called complementary angles).

∠RTC = ∠RTE + ∠ETC     Formula for complementary angles

90 = 2x + 6 + 42                Substitute the angles

90 = 2x + 48               Combined like terms to simplify

Now isolate "x" to solve

90 - 48 = 2x + 48 - 48         Subtract 48 from both sides

42 = 2x

42/2 = 2x/2       Divide both sides by 2

x = 21

Therefore x is 21.

The baby meerkat weighs 29 grams.

First, we need to ensure our units are the same, so we'll convert 776 grams to kilograms.

There are 1,000 grams in a kilogram, so 776 grams is equivalent to 0.776 kilograms.

Given that the adult meerkat is 0.747 kilograms heavier than the baby meerkat, we can calculate the baby meerkat's weight by subtracting.

Baby meerkat weight = Adult meerkat weight - Additional weight

Baby meerkat weight = 0.776 kg - 0.747 kg

Baby meerkat weight = 0.029 kg

To express this in grams, we multiply by 1,000:

Baby meerkat weight in grams = 0.029 kg × 1000

= 29 grams

The correct answer is B. I took the same test and the answer is B. :) Hope this helps!

Breann correctly solved the system of equations, combining like terms accurately to find x = 4 and y = -3.

The mistake Breann made in her calculation was combining like terms incorrectly. When she substituted y with 5 - 2x into the second equation, she arrived at x - 10 + 4x = 10. The correct combination of like terms should be 5x, not 4x, leading to the correct step 5x - 10 = 10. All other steps in her solution are correctly executed. She correctly found the value of x to be 4, and using the first equation, she correctly determined the value of y to be -3.

answer:

C. x(x^3+2)(3x-7)

hope this helps! :o)

The correct answer is D. on edg

The value of x from the given circle is 4.6 units.

From the given figure, the radius is 9.7 units.

A circle is a two-dimensional figure formed by a set of points that are at a constant or at a fixed distance (radius) from a fixed point (center) on the plane. The fixed point is called the origin or center of the circle and the fixed distance of the points from the origin is called the radius.

Here,

9.7 =5.1+x

⇒ x=9.7-5.1

⇒ x=4.6 units

Therefore, the value of x from the given circle is 4.6 units.

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To answer the questions, we need to use the concepts of probability.

To answer the questions, we need to use the concepts of probability.

a. The number of airplanes arrive and depart the airport per hour can be calculated by multiplying the probabilities of a flight departing and arriving on time. Although the exact numbers are not given, the probability of a flight departing and arriving on time is 0.81, which can be used as an estimate.

b. To find the probability that there are exactly 100 arrivals and departures in one hour, we need to use the concept of the binomial distribution. However, we don't have enough information to calculate the probability.

c. To find the probability that there are at most 100 arrivals and departures in one hour, we can use the concept of the cumulative distribution function of the binomial distribution. Again, we don't have enough information to calculate the probability.

The question is about calculating probabilities and involves understanding conditional probability, probability distributions, and statistical methods. It requires more data to answer parts a, b, and c. The remaining parts of the question can be approached with standard statistical methods and concepts.

The question pertains to the field of probability within mathematics and involves calculations based on given probabilities for airplane flights departing and arriving on time. The data presented involve conditional probabilities and the characteristics of certain probability distributions. It is essential to understand these concepts to solve the exercises provided. Unfortunately, the given information does not contain sufficient details to answer parts a, b, and c as they require additional specific data about the actual number of flights or arrivals and departures per hour. However, the remaining parts of the question can be approached with standard statistical methods such as using z-scores for normal distributions, applying properties of uniform distributions for waiting times, understanding the concept of rate-determining steps in a process, and conducting hypothesis tests about a population mean given a sample.

The answer is blue i just took the test.