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3 years ago

The lengths of the legs of a right triangle are 6 cm and 8 cm. What is the length of the hypotenuse?

2 answers:

7 0

You can find the missing side of a right triangle by using the Pythagorean Theorem (a² + b² = c²), where c is the hypotenuse. So, just plug the given numbers in and solve for c.

a² + b² = c² Plug 6 and 8 in for a and b (order doesn't matter)
6² + 8² = c² Square both numbers
36 + 64 = c² Add
100 = c² Take the square root of both sides
10 = c Switch the sides of the equation to make it easier to read
c = 10

So, the hypotenuse of the triangle is 10 cm.

Svetllana [295]3 years ago

6 0

You require the Pythagorean Theorem: in a right triangle, the sum of the squares of the legs equals the square of the hypotenuse.In this case 6^2 + 8^2 = 36 + 64 = 100 = 10^2
the hypotenuse is 10.

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Find the indicated limit, if it exists.

kondor19780726 [428]

Answer:

d) The limit does not exist

General Formulas and Concepts:

Calculus

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$
Left-Side Limit: $\displaystyle \lim_{x \to c^-} f(x)$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

Step-by-step explanation:

*Note:

In order for a limit to exist, the right-side and left-side limits must equal each other.

Step 1: Define

Identify

$\displaystyle f(x) = \left\{\begin{array}{ccc}5 - x,\ x < 5\\8,\ x = 5\\x + 3,\ x > 5\end{array}$

Step 2: Find Right-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^+} 5 - x$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} 5 - x = 5 - 5 = 0$

Step 3: Find Left-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^-} x + 3$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} x + 3 = 5 + 3 = 8$

∴ Since $\displaystyle \lim_{x \to 5^+} f(x) \neq \lim_{x \to 5^-} f(x)$ , then $\displaystyle \lim_{x \to 5} f(x) = DNE$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

5 0

2 years ago

The histogram shows the number of miles driven by a sample of automobiles in New York City. What is the maximum possible number

hjlf

Answer:

The maximum possible number of miles traveled by an automobile included in the histogram is 40,000 miles.

Step-by-step explanation:

A histogram is a bar graph reprinting the frequency of various categories.

In a histogram the y-axis represents the frequency and the x-axis represents the categories.

Consider the histogram provided for the number of miles driven by a sample of automobiles in New York City.

The y-axis represents the frequency and the x-axis represents the number of miles driven.

The category 10,000 miles has the highest frequency at 30.

And the maximum possible number of miles traveled by an automobile is 40,000 miles.

Thus, the maximum possible number of miles traveled by an automobile included in the histogram is 40,000 miles.

6 0

3 years ago

Read 2 more answers

Calculate the diameter of the circle

lianna [129]

Answer:

2.64cm

Step-by-step explanation:

A = π r^2

5.5 = π r^2

1.75 = r^2

r = 1.32

Diameter is then 2(1.32) = 2.64

3 0

3 years ago

Read 2 more answers

The level of CO2 emissions, f(x), in metric tons, from the town of Fairfax x years after they started recording is shown in the

Temka [501]

1) The average increase in the level of CO2 emissions per year from years 2 to 4 is:
Average=[f(4)-f(2)]/(4-2)=(29,172.15-26,460)/2=2,712.15/2=1,356.075 metric tons. The first is false.

2) The average increase in the level of CO2 emissions per year from years 6 to 8 is:
Average=[f(8)-f(6)]/(8-6)=(35,458.93-32,162.29)/2=3,296.64/2=1,648.32 metric tons. The second is false.

3) The average increase in the level of CO2 emissions per year from years 4 to 6 is:
Average=[f(6)-f(4)]/(6-4)=(32,162.29-29,172.15)/2=2,990.14/2=1,495.07 metric tons. The third is false.

4) The average increase in the level of CO2 emissions per year from years 8 to 10 is:
Average=[f(10)-f(8)]/(10-8)=(39,093.47-35,458.93)/2=3,634.54/2=1,817.27 metric tons. The fourth is true.

Answer: Fourth option: The average increase in the level of CO2 emissions per year from years 8 to 10 is 1,817.27 metric tons.

7 0

3 years ago

Need help please don't know this no links

abruzzese [7]

Answer:

$551

Step-by-step explanation:

The maximum profit is found where the derivative of the profit function is 0.

y' = -2x +62 = 0

x = 62/2 = 31

Then the maximum profit is ...

y = (-x +62)x -410

y = (-31 +62)(31) -410 = 961 -410 = 551

Assuming the profit is in dollars, the maximum profit the company can make is 551 dollars.

4 0

2 years ago