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2 years ago

Which value of x would make Triangle S U V Is-congruent-to Triangle T U M by HL?

Mathematics

2 answers:

Katarina [22]2 years ago

6 0

Answer:

$x=5$

Step-by-step explanation:

Actually, the complete question includes the expression of each hypothenuse, because the problem is about congruence between right triangle.

So, the hypothenuse for $\triangle SUV$ is: $2x+9$ .

The hypothenuse for $\triangle TUW$ is: $4x-1$ .

So, the postulate HL (Hypothenuse-Leg) states that both right triangles are equal if their hypothenuses and legs are equal. So, we have to make equal both hypothenuse expression and solve for $x$ :

$2x+9=4x-1\\9+1=4x-2x\\10=2x\\x=\frac{10}{2}=5$

Therefore, when $x=5$ these triangles are congruent.

Vaselesa [24]2 years ago

3 0

Answer:

Step-by-step explanation:

2x + 9 = 4x -1

1. substract the 2x from both sides so you end up with 4x - 2x = 2x

2. add the one to both sides so 9+1= 10

3. then divide 10 and 2x = 5

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A tank contains 8000 L of pure water. Brine that contains 35 g of salt per liter of water is pumped into the tank at a rate of 2

OleMash [197]

Answer:

The concentration of salt in the tank approaches $35 \mathrm{g} / \mathrm{L},$

Step-by-step explanation:

Data provide in the question:

Water contained in the tank = 8000 L

Salt per litre contained in Brine = 35 g/L

Rate of pumping water into the tank = 25 L/min

Concentration of salt $\lim _{t \rightarrow \infty} C(t)=\lim _{t \rightarrow \infty} \frac{35 t}{320+t}$

Now,

Dividing both numerator and denominator by $t$ , we have

$\lim _{t \rightarrow \infty} \frac{\frac{1}{t} 35 t}{\frac{1}{t}(320+t)}=\lim _{t \rightarrow \infty} \frac{35}{\frac{320}{t}+1}=35$

Here,

The concentration of salt in the tank approaches $35 \mathrm{g} / \mathrm{L},$

3 0

2 years ago

Flip two coins 100 times, and record the results of each coin toss in a table like the one below:

monitta

Answer:

1)The theoretical probability that a coin toss results in two heads showing is 25%.

2)The experimental probability that a coin toss results in two heads showing is 44%.

3) The theoretical probability that a coin toss results in two tails showing is 25%.

4) The experimental probability that a coin toss results in two tails showing is 34%.

5) The theoretical probability that a coin toss results in one head and one tail showing is 50%.

6) The experimental probability that a coin toss results in a head and a tail is 22%.

7) The experimental probabilities are slightly different from the theoretical probabilities because the number of experiments is relatively small. As the number of experiments increase, the experimental probabilities will get closer to the theoretical probabilities.

Step-by-step explanation:

Probability:

What you want to happen is the desired outcome.

Everything that can happen iis the total outcomes.

The probability is the division of the number of possible outcomes by the number of total outcomes.

Theoretical Probability:

The results you expect to happen.

Experimental Probability:

The probability determined from the result of an experiment.

1. What is the theoretical probability that a coin toss results in two heads showing?

In each toss, the theoretical probability that a coin toss results in a head showing is 50%.

So for two coins, the probability is:

$P = (0.5)^{2} = 0.25$

The theoretical probability that a coin toss results in two heads showing is 25%.

2. What is the experimental probability that a coin toss results in two heads showing?

There were 100 flips, and it resulted in two heads 44 times, so:

$P = \frac{44}{100} = 0.44$

The experimental probability that a coin toss results in two heads showing is 44%.

3. What is the theoretical probability that a coin toss results in two tails showing?

In each toss, the theoretical probability that a coin toss results in a tail showing is 50%.

So for two tails, the probability is:

$P = (0.5)^{2} = 0.25$

The theoretical probability that a coin toss results in two tails showing is 25%.

4. What is the experimental probability that a coin toss results in two tails showing?

There were 100 flips, and it resulted in two tails 34 times, so:

$P = \frac{34}{100} = 0.34$

The experimental probability that a coin toss results in two tails showing is 34%.

5. What is the theoretical probability that a coin toss results in one head and one tail showing?

In each toss, the theoretical probability that a coin toss results in a tail showing is 50% and in a head showing is 50%.

They can be permutated, as the tail can appear before the head, or the head before the tail. So:

$P = p_{2,1}*(0.5)*(0.5) = \frac{2!}{1!}*0.25 = 0.50$

The theoretical probability that a coin toss results in one head and one tail showing is 50%.

6. What is the experimental probability that a coin toss results in one head and one tail showing?

There were 100 flips, and it resulted in a head and a tail showing 22 times, so:

$P = \frac{22}{100} = 0.22$

The experimental probability that a coin toss results in a head and a tail is 22%.

6 0

2 years ago

Please help!!!!!!!!!

shepuryov [24]

A, the first one only, this parabola only has a minimum and no maximum. the other statements are also just false

7 0

2 years ago

Xenia has 100,000,000 people. Of this population, 25,000,000 residents are below age 16, and 10.000,000 have given up looking fo

Goryan [66]

Answer: 45000

Step-by-step explanation:

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2 years ago

You are researching the penguin population in the Falkland Islands. There are 1000 penguins in the beginning of the first year:

kolbaska11 [484]

Answer:

The formula to determine the population of penguins at the end of the 7th year is:

$P_7=1.3^7\cdot P_0=6.275\cdot 1000=6275$

Step-by-step explanation:

With this information, we know that the initial population at end of year 0 is 1000 penguins.

The first year will born 500 chicks (50% of the population) and also 20% of the total population will die.

We can then model the population for the end of year 1 as:

$P_1=P_0+0.5P_0-0.2P_0=1.3P_0$

As this dynamic will continue with the years, we can generalize as:

$P_2=1.3P_1=1.3\cdot (1.3P_0)=1.3^2P_0\\\\P_n=1.3^nP_0$

Then, the value of the population at the end of the 7th year should be:

$P_7=1.3^7\cdot P_0=6.275\cdot 1000=6275$

8 0

2 years ago